Number of results: 121,545
Find the heat necessary to change 2.5kg of water (ice) at -2 degrees C to steam at 105 degrees C. specific heat ice= 2060 specific heat water= 4180 specific heat steam= 2020 heat of fusion= 3.34X10^5 heat of vaporization= 2.26X10^6
Thursday, November 18, 2010 at 8:49pm by Brittany
heat=force*distance=30*12*.075 Joules heat=mass*specificheat*DeltaTemp now for the specific heat of skin, it is mostly water, so I would use the specific heat of water for that. Solve for delta Temp
Friday, July 19, 2013 at 7:01pm by bobpursley
q = mass x specific heat x delta T. 176 J = 28.1 x specific heat x 16.3 Solve for specific heat and compare with the table provided.
Sunday, January 26, 2014 at 7:56pm by DrBob222
Chem. Find specific heat capacity.
[massPb x specific heat Pb x (Tfinal-Tinitial)] + [massH2O x specific heat H2O x (Tfinal-Tinitial)] = 0 Only one unknown; i.e., specific heat Pb. Solve for that. Post your work if you get stuck.
Sunday, November 30, 2008 at 8:25pm by DrBob222
Note the correct spelling of celsius. q = mass x specific heat water x (Tfinal-Tinitial) For calories specific heat = 1 cal/g*C For joules specific heat = 4.184 J/g*C
Sunday, April 3, 2011 at 7:06pm by DrBob222
q = mass x specific heat x delta T Solve for specific heat. I think Cu has a specific heat of about 0386 J/g*C.
Tuesday, June 7, 2011 at 8:25am by DrBob222
heat capacity has units of J/degree C. specific heat has units of J/g*C So specific heat x grams = heat capacity. Plug in g and heat capacity and solve for specific heat. I get something like 0.2 J/g*c
Wednesday, January 20, 2010 at 10:43pm by DrBob222
specific heat is a shortened version of specific heat capacity. Use 1 = mass x specific heat x delta T.
Sunday, June 6, 2010 at 7:45pm by DrBob222
heat lost by metal + heat gained by water = 0. [mass metal x specific heat M x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0 Substitute and solve for specific heat. Then look in your text or on-line for a list of specific heats of metals and ...
Saturday, May 14, 2011 at 10:36pm by DrBob222
heat lost by metal + heat gained by H2O = 0 [mass metal x specific heat metal x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0 Substitute and solve for specific heat metal, then look in specific heat tables and identify the metal with that specific...
Friday, March 15, 2013 at 2:56pm by DrBob222
Physics - Specific heat
#1.) If 500 grams of a metal requires 70 calories of energy to raise its temperature by 10 Kelvin, what is its specific heat? #2.) How much heat is gained or lost when 200 grams of ice melts?
Thursday, October 28, 2010 at 12:24am by Heath
how much heat is required to completely vapporize 2.3 grams of ice starting at -50 degrees Celsius? (the specific heat of ice is 2.0 J/g.C; the specific heat of water is 4.184 J/g.C; the heat of fusion is 333 J/g; and the heat of vaporization is 22601 J/g)
Tuesday, March 22, 2011 at 12:04am by Anonymous
A 100-g aluminum calorimeter contains a mixture of 40 g of ice and 200 g of water at equilibrium. A copper cylinder of mass 300 g is heated to 350 C and then dropped into the calorimeter. What is the final temperature of the calorimeter and its contents if no heat is lost to ...
Thursday, December 2, 2010 at 5:22pm by Calvin
q = mass x specific heat x (Tfinal-Tinitial) Note that you list the specific heat Fe but ask for how much heat is required to heat Al. You may have mixed two problems.
Wednesday, April 4, 2012 at 1:26pm by DrBob222
metal loss of heat + water gain of heat = 0 [(mass metal x specific heat metal x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0 Substitute the numbers and solve for specific heat metal. Then look up in a specific heat table to identify the metal.
Monday, March 18, 2013 at 2:59pm by DrBob222
Please show some effort of your own other than changing a letter in your name each time you post. (1) Look up the specific heats of glass, aluminum and water. (2) Compute the amount of heat energy gained by those three materials as they heat from 12.5 to 35 C. Call that heat Q...
Thursday, May 6, 2010 at 12:38am by drwls
Heat gained = Heat loss Let final temperature be x, Specific heat of Iron: 0.12 cal/g°C = 502 J/kg °C Specific Heat of Water: 4186 J/kg °C Heat gained by water = mass x specific heat capacity x change in temperature =100(4186)* (x-25) =(418600x-10465000)J Heat lost by nails = ...
Tuesday, April 3, 2012 at 3:42pm by Cecilia
Q = heat gained by silver = (Specific Heat of silver)*5g * 79 C = ? The specific heat of silver is 0.237 J/gC Q = 93.6 J = 22.4 calories
Monday, January 31, 2011 at 2:50am by drwls
q = mass x specific heat x delta T. 6.1 = 1.50g x specific heat x 9 Solve for specific heat.
Monday, January 16, 2012 at 9:44pm by DrBob222
the heat of fusion water is335 J/g. The heat of evaporization of water is 2.26 kJ/g, the specific heat of ice is 2.05 J/Deg/g, the specific heat of steam is 2.08 J/deg/g and the specific heat of liquid water is 4.184 J/deg/g. How much heat would be needed to convert 12.09 g of...
Tuesday, April 24, 2012 at 10:31pm by Randy
Power*time= Heat= Mass*Hvwater + Mass*(100-21)*specific heat of water. Watch specific heat, heat of vaporization units, make certain they match your mass units. Solve for time.
Sunday, January 20, 2008 at 9:07pm by bobpursley
ocean county college
A 150 g sample of metal at 55.0 C raises the temperature of 150 g of water from 23.0 C to 35.0 C. Determine the specific heat of the metal in J/g C. Hint: #1 First find heat absorbed by water using formula q = mass x specific heat x change in temp (of water). #2 Use this value...
Sunday, March 21, 2010 at 8:10pm by adina
q = mass x specific heat x (Trinal-Tinitial) q = 550 x specific heat Al x (145-22) Look up the specific heat Al in your text or notes and punch the calculator.
Monday, April 11, 2011 at 7:16pm by DrBob222
look up the specific heat capacity for lead. Heat= mass*specificheat*(Tf-ti) make certain mass is in the same units as the specific heat capacity is. Why do you keep posting this question?
Monday, December 12, 2011 at 10:26am by bobpursley
3. Based on your answer in question 2, determine the percent error of your calculated specific heat capacity value of your unknown metal using the specific heat capacities of known metals below. Metal Specific Heat Capacity Nickel 0.440 Tin 0.210 Silver 0.237 Magnesium 0.140 ...
Thursday, April 4, 2013 at 2:35pm by lex
heat lost by alloy + heat gained by water = 0 heat lost or gained = mass x specific heat x (Tfinal-Tinitial). mass alloy x specific heat alloy x (Tfinal-Tinitial) + mass water x specific heat water x (Tfinal-Tinitial) = 0 You must look up the specific heat of water, the ...
Wednesday, June 11, 2008 at 10:01am by DrBob222
Integrated Physics and Chemistry
Think about the equation that defines specific heat: Q = M*C*(delta T) In your case, Q (the added heat) and M (the mass) are the same for gold and iron. That means delta-T (the change in termperature) is inversely proportional to C, the specific heat. delta T = (Q/M)*(1/C) You...
Monday, March 21, 2011 at 11:18am by drwls
calculate the specific heat of a metallic element if 314 joules of energy are needed to raise the teperature of a 50.0g sample from 25C to 50C. i'm not sure if you use the Q=(s)(m)(delta t) equation or not since it asks for the specific heat not the specific heat capacity.. i ...
Sunday, June 6, 2010 at 7:45pm by Ashley
Physics help please
Heat lost by the coins + heat gained by the water = 0 mass x specific heat x (Tf - Ti) + mass x specific heat x (Tf - Ti) = 0 Tf is final T. Ti is initial T. Post your work if you get stuck.
Tuesday, December 4, 2007 at 11:54am by DrBob222
you have three parts: a. Heat the water to 100C b. vaporize the water at 100C c. heat the steam to 110C a. heat=m*cwater*(100-20) b. heat=m*Hv c. heat=m*csteam*(110-100) cwater is specific heat capacity of water Hv is the heat of vaporization of water. Csteam is specific heat ...
Sunday, March 31, 2013 at 6:28pm by bobpursley
The amount of heat required to raise temperature of 55.85 g of iron 1 degrees C is called its A. Change of energy B. Enthalpy C. Molar heat capacity D. Specific heat capacity E. Specific Heat.
Friday, November 16, 2007 at 4:42pm by Chemistry
(g steam x heat vap) + (mass ice x heat fusion) + [(mass steam H2O x specific heat x (Tfinal-Tinitial)] + [mass melted ice x specific heat x (Tfinal-Tinitial)]= 0 I don't know what units you ae using; you substitute them. (1g x -heat vap) + (xgrams x heat fusion) + [(1g x ...
Sunday, March 10, 2013 at 11:03pm by DrBob222
I'm understanding most of this physics, but here is one last question I would appreciate help on. Thank you! Okay, so the question involves a figure. A link to the question and the picture with it is below can be found by googling the question along with Holt physics. The link...
Friday, March 26, 2010 at 2:42pm by Demi
No. You have to use units here. What units of mass, specific heat did you use? Notice the units in the table for specific heat, and heat vaporization. That is where You erred.
Sunday, March 31, 2013 at 6:28pm by bobpursley
heat = masstotal*specificheatwater*6.5 = .050kg*2*6.5C*specific heat so look up the specific heat of water in units of joules/kg-C
Sunday, April 8, 2012 at 9:46pm by bobpursley
I'm surprised they aren't given in the problem. At any rate, look up the specific heat of Al from problem 1 and Cd from problem 2. Problem 3 gives you temperatures and you are to calculate specific heat. The specific heat of water is 4.184 Joules/grams or 1.0 calories/gram.
Wednesday, September 23, 2009 at 8:18pm by DrBob222
heat removed from water in going from 21.0 to zero C is mass x specific heat x delta T. mass = 33.6 g specific heat is 4.184 J/g*K delta T is 21.0 - 0. heat added to ice to melt it is mass x heat fusion. mass is the unknown. heat fusion is given. Note: you need to work in the ...
Monday, February 4, 2008 at 12:52pm by DrBob222
q = mass x specific heat x delta T. q = the heat absorbed. mass = mass of the water. That is listed in the formula. 1.0 L H2O has a mass of 1000 g. specific heat. Do you know the specific heat of water. It goes here. delta T is the difference in temperature, in this case it is...
Wednesday, May 19, 2010 at 10:23pm by DrBob222
a 3.55 kg sample of an alloy is heated to 295 degree and plunged into an aluminium vessel containing 10.0L of water. the initial temperature of the vessel and water is 19 degree.the mass of the aluminium vessel is 2.1 kg and the specific heat of aluminium is 900 J kg-1K-1. if ...
Thursday, April 26, 2007 at 5:01am by Twinkle
q = mass x specific heat x (Tfinal-Tinital) Substitute and solve for the only unknown in the equation; i.e., specific heat. It releases heat; therefore, q is -12.5 cal.
Tuesday, April 16, 2013 at 7:26pm by DrBob222
What would happen to the specific heat if some of the warm metal shot were lost during the transfer to the calorimeter? Answer a. It would not affect the calculated value of specific heat b. It would cause the calculated value of specific heat to be artificially high c. It ...
Sunday, October 24, 2010 at 11:33pm by kellie
call specific heat of gold G call specific heat of water W (I am not about to get a book out) 3 (G) (99-T) = .22 (W)(T-25) T will be pretty close to boiling.
Wednesday, January 25, 2012 at 11:11pm by Damon
You have it a little mixed up. You don't have specific heat metal at all; you have specific heat H2O added instead of multiplied. heat lost by metal + heat gained by H2O = 0 [mass metal x specific heat metal x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)
Monday, March 18, 2013 at 4:57pm by DrBob222
Call final temp = T Heat lost by Al = specific heat AL * mass of Al * (80-T) Heat gained by water - specific heat of H2O * mass of H2O *(T-20) Then Heat lost by Al = Heat gained by H2O
Sunday, July 4, 2010 at 8:57am by Damon
heat=mass*specific heat*deltaTemp look up the specific heat of Hg.
Sunday, October 5, 2008 at 12:56pm by bobpursley
Heat lost by spoon = heat gained by water Use that heat, the aluminum temperature drop, and the specific heat of aluminum to get the mass. Ignore the styrafoam mass. It acts as an insulator to keep the heat inside, and weighs very little.
Saturday, May 7, 2011 at 12:56am by drwls
If I'm not mistaken, it's 9.097155832 grams, which you can round as you see fit. 913.5 J = m x specific heat water x 24 degrees Celcius. Specific heat of water is 4.184 J/(gC). Solve for m.
Saturday, November 22, 2008 at 11:05pm by Jeremy
The rod is cooler than the water; therefore, the rod will gain heat and the water will lose heat and the two added together will be zero (one will be negative). massrod x specific heat rod x (Tfinal-Tinitial) + massH2O x specific heat water x (Tfinal-Tinitial) = 0 Solve for ...
Tuesday, October 28, 2008 at 4:54pm by DrBob222
All right. I'll go try and work on this problem and come back if I need some help. The specific heat of ice: 2.092 J/g-Kelvin The specific heat of water: 4.184 J/g-Kelvin The specific heat of steam: 1.841 J/g-Kelvin Heat of Fusion: 6.008 kJ/mol at 0.00 degrees Celsius Heat of ...
Saturday, December 11, 2010 at 8:50pm by Tichele
First convert 175 F to C. I think that's about79 but you need to do it more accurately. You have omitted one piece of info; i.e., the specific heat of solid H2O(ice). heat lost by ice in going from -18C to zero C + heat lost by ice melting at zero C + heat gained by melted ice...
Friday, January 11, 2013 at 10:49pm by DrBob222
Q = M*C*(delta T), where Q is the heat removal required M = mass of ice C = specific heat of solid ice delta T = temperature change (-30 C in this case). You will need to look up the specific heat of solid ice.
Sunday, August 14, 2011 at 3:43pm by drwls
[mass Cu x specific heat Cu x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0 Solve for specific heat Cu. It's easier to put in all the numbers first into the above, do the math, then solve for specific heat Cu last. Post your work if you get stuck...
Friday, January 2, 2009 at 7:52pm by DrBob222
heat lost by hot metal + heat gained by cooler water = 0 [mass metal x specific heat metal x (Tfinal-Tinitial)] + [mass H2O x specific heat water x (Tfinal-Tinitial)] = 0 Solve for specific heat metal. You have all of the other items.
Saturday, January 21, 2012 at 9:27pm by DrBob222
Do you have the heat of fusion? The specific heat of ice? The specific heat of H2O?
Saturday, December 8, 2012 at 12:26am by DrBob222
You need to raise the temperature of the lead from 27.3 degrees C to 327.3 degrees C heat required = mass * specific heat of lead * (327.3-27.3) then melt the lead heat to melt = mass * heat of fusion of lead You know everything here but the specific heat of lead which you ...
Tuesday, March 30, 2010 at 10:15pm by Damon
Heat loss from M grams steam cooling and condensing to 50 C liquid = heat gained by 250 g water increasing from 22 to 50 C PLUS heat gained by beacker. Write that as an equation and solve for M. You will need to know the specific heat of steam in the gas phase (about 0.4 cal/g...
Thursday, January 20, 2011 at 4:54pm by drwls
What's high specific heat is mainly a consequence of the: A) high specific heat of oxygen and hydrogen atoms B) inability of water to dissipate heat into dry air D) absorption and release of heat when hydrogen bonds break and form E) fact that water is a poor heat conductor I ...
Thursday, October 18, 2007 at 9:26pm by J
I don't know of a site. Two things you need to know. heat lost + heat gained = 0 and mass x specific heat x delta T = heat lost or gained by anything. so. heat lost by Al + heat lost by Fe + heat gained by water = 0 heat lost by Al = mass x specific heat x (Tf - Ti) where Tf ...
Tuesday, November 6, 2007 at 11:22pm by DrBob222
An unknown substance is found to require a very large amount of heat energy to raise its temperature a single degree. Does it have a high specific heat capacity or a samall specific capacity? Explain
Saturday, March 3, 2012 at 3:48pm by GRISSELL
The potential energy becomes heat, Q. Use the heat to compute the temperature rise. Delta T = Q/(M*C) = (1/2)M g H/(M*C) = (1/2)g H/C C is the water specific heat, 4180 joules/kg C
Tuesday, June 28, 2011 at 12:39am by drwls
heat lost by metal + heat gained by water = 0 [mass metal x specific heat metal x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0 Substitute the numbers and solve for the only unknown, the specific heat of the metal.
Monday, April 16, 2012 at 12:26am by DrBob222
q1 = heat absorbed by water = mass H2O x specific heat H2O x delta T. For part 2, I believe it's best to do it all in one. heat lost by Cu + heat gained by water = 0 heat lost by Cu = mass Cu x specific heat Cu x delta T. heat gained by water = done in part 1. one unknown, ...
Sunday, April 20, 2008 at 11:33am by DrBob222
specific heat of Aluminom
What is the specific heat of Aluminom if it takes 1100 calories to take the temp. from 10 degrees C to 80 degrees C and the block of Aluminon has a mass of 60 grams? q = massAl x specific heat x (Tf-Ti) q is 1100 calories Tf is final T Ti is initial T mass is 60g solve for ...
Tuesday, December 19, 2006 at 9:21pm by brandi
Heat lost by lead = heat gained by water. Write that with T(final) as the only unknown, using masses and specific heats, and solve for T(final), the equilibrium temperature. You will need to look up the lead specific heat. I am sure you know the value for water. None of the ...
Saturday, May 7, 2011 at 12:37am by drwls
I have also assumed that the "gamma" (specific heat ratio Cp/Cv) is the same for air at both temperatures (27 and 127 C). This is a valid assumption up to about 200 C. http://www.engineeringtoolbox.com/specific-heat-ratio-d_602.html speed of sound = sqrt(gamma*R*T/M) M = ...
Tuesday, December 25, 2012 at 4:16am by drwls
please list all the formulas for calculating the specific heat of a substance please. thanks specific heat = (heat added)/[(mass)(delta T)] That is the equation that defines specific heat. It should be the only equation you need. "delta T" is the increase in temperature. For ...
Sunday, April 8, 2007 at 2:52am by Pauline
The heat of fusion of water is 335 J/g, the heat of vaporization of water is 2.26 kJ/g, the specifc heat of ice is 2.05 J/deg/g, the specific heat of steam is 2.08 J/deg/g and the specific heat of liquid water is 4.184 J/deg/g. How much heat would be needed to convert 10.73 g ...
Wednesday, May 8, 2013 at 12:41pm by Jenny
The water will be heated by each metal. How much. q = mass H2O x specific heat H2O x delta T. Therefore for dT to be the smallest, you want q to be the smallest How much will the metal change? q = mass metal x specific heat metal x (Tfinal-Tinitial) q = 10g x specific heat ...
Wednesday, May 1, 2013 at 12:23am by DrBob222
You must do the (a) part in stages. q1 = heat required to move the ice from a T of -5 to zero. mass x specific heat ice x (Tfinal-Tinitial). You have mass, look up specific heat ice and delta T = 5. q2 = heat to melt the ice. q2 = mass x heat fusion. You have mass. Look up ...
Saturday, November 22, 2008 at 10:06pm by DrBob222
The initial heat of the water and can is 600*Cw*25 + 55*0.1*25 + x*Cw*65 The final heat of the water and can is (600+x)*Cw*31 + 55*0.1*31 Set the inital and final heat equal to each other, and solve for x; You will have to look up the specific heat of water (Cw), it should be ...
Wednesday, December 5, 2012 at 8:26pm by Jennifer
heat gained by first substance + heat lost by water = 0 [mass*specific heat x (Tfinal-Tinitial)] + [massH2O x specific heat water x (Tfinal-Tnitial)] = 0 Solve for Tfinal. Post your work if you get stuck.
Thursday, February 24, 2011 at 2:02pm by DrBob222
A 50.0 g silver spoon at 20.0 degrees Celsius is placed in a cup of coffee at 90.0 degrees C. How much heat does the spoon absorb from the coffee to reach a temp. of 89.0 degrees C? Rememeber this equation. It, or a slight modification of it, will work all of these heat ...
Tuesday, November 21, 2006 at 4:36pm by Erica
The heat lost by the aluminum equals the heat gained by the water. Look up the specific heat of aluminum, C_a, and write the above statement as an equation. Let C_w be the specific heat of water. M_a*C_a*(100 - T) = M_w*C_w*(T - 20) M_a = 41 g M_w = 71 g C_w = 1.0 cal/C*g ...
Monday, April 25, 2011 at 10:30pm by drwls
The specific heat of water is 4.184 J/g degree C, and the specific heat of iron is 0.451 J/g degree C, which will reach a higher temperature when exposed to a given amount of heat?
Tuesday, October 2, 2012 at 8:06pm by Alz
Calculate the heat released when 64.5g of steam at 119 C is converted to water at 36 C. Assume the specific heat is 4.184 J/g*C for water, the specific heat of steam is 1.99J/g*C, and the heat of vaporization is 40.79 kJ/mol for water.
Friday, June 17, 2011 at 6:59pm by Sydney
q = mass x specific heat x delta T. I assume 40.0 blank is 40.0 grams silver. q will be in joules if you use specific heat in J/g*C and it will be in calories if you use specific heat in calories/g*C.
Monday, June 14, 2010 at 1:09pm by DrBob222
You need more information. You need the specific heat of the glass in order to calculate the specific heat of the liquid. The information you provided will only tell you the ratio of speciic heats
Monday, December 13, 2010 at 11:46pm by drwls
heat lost by unknown compound + heat gained by H2O = 0 [mass unknown x specific heat unk x (Tfinal-Tinitial)] +[mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0 Substitute and solve for specific heat unknown.
Monday, March 25, 2013 at 8:04pm by DrBob222
what is the relationship between density and specific heat capacity? and why? i thought when density increase specific heat increase but it says that it is inversely proportional. thank you
Wednesday, December 1, 2010 at 11:16pm by lizzy
d question is not complete.either d (specific heat capacity of ice) is give or(specific heat of fusion of ice)is given so check d question.
Thursday, March 17, 2011 at 9:49pm by pete onoriode
50 * (80-45)(specific heat capacity of water) specific heat capacity of water is about 4190 Joules/kg degK
Wednesday, January 18, 2012 at 5:55pm by Damon
20 kg [ specific heat of ice in cals or joules /kg deg * 6 + heat of fusion of water + specific heat of water * 20 ]=answer in calories or Joules
Wednesday, January 27, 2010 at 6:20pm by Damon
Which holds the most heat? q = mass metal x specific heat x delta T. q = 10g x sp.h. x (100-25). To make q large, specific heat must bae (large/small)?
Monday, November 7, 2011 at 3:23pm by DrBob222
Convert MeV to Joules/proton and multiply that by 5.5*10^16 protons/s That gives you the answer in watts (or J/s). Watts divided by (mass * specific heat) is the rate of temperature increase You will need to look up the specific heat of copper.
Sunday, February 7, 2010 at 12:02am by drwls
How much heat is required to raise the temperature of a 10.0 kg of aluminum by 80 degrees Celsius? Look up the specific heat of aluminum. The units will be Calories/kg*degC or Joules/kg*degC. Multiply that number by the temperature change (80 C) and the mass (10.0 kg). See ...
Friday, June 22, 2007 at 8:29pm by jan
I think you are right but I like this way of doing it better. loss of heat from substance + heat gained by water = 0 [mass substance x specific heat substance x (Tfinal-Tinitial)] + [mass water x specific heat water x (Tfinal-Tinitial) = 0 Solve for specific heat substance.
Monday, March 8, 2010 at 4:25pm by DrBob222
q1 = heat to warm ice from -10 to zero degrees C (without melting it). q1 = mass ice x specific heat ice x delta T (delta T is 10). q2 = heat to melt ice at zero degrees C. q2 = mass ice x heat fusion. q3 = heat to raise temperature of water zero C to +40. q3 = mass water x ...
Sunday, April 13, 2008 at 5:56pm by DrBob222
heat lost by metal + heat gained by H2O=0. [mass metal x specific heat metal (Tfinal-Tinitial)] + [mass water x specific heat water x (Tfinal-Tinitial)] = 0 Solve for specific heat metal, the only unknown..
Sunday, October 31, 2010 at 5:12pm by DrBob222
heat lost by Fe + heat gained by H2O = 0 [mass Fe x specific heat Fe x (Tfinal-Tinitial)] + [mass H2O x specific heat water x (Tfinal-Tinitial)]= 0 Substitute and solve for specific heat Fe.
Sunday, May 27, 2012 at 9:41pm by DrBob222
A quantity of ice at 0.0 degrees C was added to 33.6 of water at 21.0 degree C to give water at 0.0 degrees C. How much ice was added? The heat of fusion of water is 6.01 kJ/mol and the specific heat is 4.18 J/(g * degrees C) q = mass x specific heat x delta T? q = mass x heat...
Monday, February 4, 2008 at 4:42pm by Lauren
Do you know the specific heat of solid water (ice)? If so, then it will take q calories = mass ice x specific heat ice x (0+10) to raise T from -10 C to zero C. Subtract that from 4500 to see how much(if any) heat is left. If some heat is available, then it can go to melting ...
Tuesday, January 3, 2012 at 3:20pm by DrBob222
I can't read your equation but I now what you want. There are two equations you need to do this. The heat transfer WITHIN a phase is q = mass x specific heat in that phase x (Tfinal-Tinitial). The second equation you need is for heat transfer AT THE PHASE CHANGE. q = mass x ...
Friday, November 2, 2012 at 4:30am by DrBob222
An ice chest at a beach party contains 12 cans of soda at 2.45 °C. Each can of soda has a mass of 0.35 kg and a specific heat capacity of 3800 J/(kg C°). Someone adds a 9.60-kg watermelon at 25.3 °C to the chest. The specific heat capacity of watermelon is nearly the same as ...
Sunday, May 5, 2013 at 12:47pm by Tim
An ice chest at a beach party contains 12 cans of soda at 4.16 °C. Each can of soda has a mass of 0.35 kg and a specific heat capacity of 3800 J/(kg C°). Someone adds a 8.48-kg watermelon at 25.7 °C to the chest. The specific heat capacity of watermelon is nearly the same as ...
Friday, February 7, 2014 at 4:13pm by Nicj
Chemistry 106 ch 11
Calculate the heat released when 73.5 g of steam at 122.0 C is converted to water at 39.0C. Assume that the specific heat of water is 4.184 J/g C, the specific heat of steam is 1.99 J/g C, and change in heat vap =40.79 kJ/mol for water.
Saturday, February 15, 2014 at 6:23pm by alex
heat lost by alloy + heat gained by water + heat gained by coffee cup = 0 You have three segments above. Substitute the following into the appropriate part and solve for the only unknown in the bunch which is specific heat of the alloy. Heat lost by alloy is as follows: mass ...
Friday, October 18, 2013 at 11:03pm by DrBob222
heattoheatwater=masswater*specific*(45-20) figure mass water from volume and denstity, look up specific heat. if eff is .98, then energy=heat/.98 power=heateneryg/timeinseconds R=V^2/energy
Sunday, December 5, 2010 at 6:33pm by bobpursley
Heat a known mass of a metal to a known temperature (say, for example, 10.00 g at 78.5 deg C) Add the metal to a known mass of water at a known temperature (say 75.00 g at 25.0 deg C) in a calorimeter. Let the metal and the water come to an equilibrium temperature, Tf. (The ...
Tuesday, April 16, 2013 at 10:48pm by Rebekah
science so confused??
heat capacity; temperature ; equal; calorimeter. Now on the first, the question is poorly worded. Heat given off is equal to the product of mass x specific heat capacity x change in temperature. Those three things define heat change. In olden days, mass x specific heat content...
Tuesday, April 22, 2008 at 1:18am by bobpursley
the specific heat capacity of gold is .13 J/g°C. Calculate the specific heat capacity of gold in cal/g°C. Calculate the specific heat capacity of gold in J/mol°C. Calculate the amount of energy required in joules to heat 2.5 kg of water from 18.5°C to 55.0°C I think for these ...
Monday, May 31, 2010 at 11:03am by Sheena