Wednesday

April 23, 2014

April 23, 2014

Number of results: 2,158

**Integrated Physics and Chemistry**

The specific heat capacity of copper is 0.09 cal/g°C. How much energy is needed to flow into a 10-gram sample to change its temperature from 20°C to 21°C? A)0.009 cal B)0.09 cal C)0.9 cal D)9 cal
*Thursday, April 21, 2011 at 10:29am by Ashley!!*

**math**

80 cal/g x 5 g 400g/cal or 400 cal/g not sure can any one help Neither. 80 cal/g x 5g = 400 cal. The unit g cancels to leave calories. x2+2x-15
*Monday, December 11, 2006 at 10:45pm by rena*

**chemistry**

How many calories are required to heat 25.0 g of platinum from 24.5 °C to 75°C (specific heat of platinum = 0.139 J/gK)? 1. 47 cal 2. 42 cal 3. 48 cal 4. 20 cal 5. 80 cal Thank you.
*Sunday, November 22, 2009 at 5:30pm by Danny*

**science**

the conversion factor for calories (cal) and joules (J) is: 1 cal = 4.184 J now to convert calories to joules, we multiply 4000 cal by 4.184 J / 1 cal so that the cal unit will be canceled, and the numerator will have the unit J: 4000 cal * 4.184 J / cal = 16736 J hope this ...
*Wednesday, October 26, 2011 at 3:09am by Jai*

**Science**

A 50 gram ice cube is cooled to -10oC in the freezer. How many calories of heat are required to heat it until it becomes liquid water at 20oC? (The specific heat of ice is 0.5 cal/goC.) Answer A) 1250 cal. B) 4000 cal. C) 5000 cal. D) 5250 cal. Can someone please help me solve...
*Monday, September 26, 2011 at 1:29am by Anonymous*

**physics **

A 50 gram ice cube is cooled to -10oC in the freezer. How many calories of heat are required to heat it until it becomes liquid water at 20oC? (The specific heat of ice is 0.5 cal/goC.) Answer A) 1250 cal. B) 4000 cal. C) 5000 cal. D) 5250 cal. Can someone please help me solve...
*Monday, September 26, 2011 at 5:17pm by Allison*

**Chemistry**

You don't need the heat of vaporization to do that calculation. It takes 10g*1Cal/C g*10C = 100 Cal heat release to reduce the temperature ot 0C, and another 80 Cal/g*10 g = 800 Cal to freeze it. The sum is 900 Cal
*Saturday, April 17, 2010 at 12:35am by drwls*

**college maths**

Mathematically, you can set up the problem as follow: 232 Cal/1.5 Cup = x Cal/1 cup 1x = 232/1.5 x = 154.7 Cal Another way, this is a no brainer question given the choices below. With 1.5 cup, you know it has 232 Cal, so with only 1 cup, you know the answer should be less than...
*Wednesday, December 15, 2010 at 2:16pm by Terry*

**Chemistry**

Do you know the conversion factor? 1 cal = 4.184 joules and I get 4565 cal. Is that a capital C for cal. In nutrition, especially in the old days, there was a big calorie (actually a kilocalorie or Calorie) and a small calorie (a calorie). If we round the 5 to the even number...
*Friday, October 30, 2009 at 7:22pm by DrBob222*

**Chemistry**

1.24E7 J x (1 cal/4.184J) = 2.98E6 cal. 12 oz x 2.98E6 cal/80,000 cal = ?oz Gatorade.
*Saturday, April 20, 2013 at 7:55pm by DrBob222*

**physics**

In cooling down to liquid at 0 C, the steam will lose Q = 540 cal/g*15g + 1.0cal/gC*100C*15g = 8100 + 1500 = 9600 calories. That amount of heat, transferred to the ice, can melt 9600 cal/80 cal/g = 120 g of ice. The 540 cal/g and 80 cal/g numbers that I used are the heats of ...
*Tuesday, April 3, 2012 at 9:25pm by drwls*

**Chemistry**

How much heat is released when 10 g of steam at 100°C is cooled to 10 g of liquid water at 100°C? I am working on a practice test and the answer is not in the back of the book. Can some one help me. 1. 10 cal 2. 5400 cal 3. 800 cal 4. 6200 cal
*Thursday, November 10, 2011 at 2:40am by nan*

**chemistry**

334 J/g x (1 cal/4.184 J) = ? cal/g. Note how i knew that the factor is 1 ca/4.184 J and not 4.184 J/1 cal. The unit we don't want cancels(J); the unit we want to keep stays(cal).
*Thursday, June 20, 2013 at 6:04pm by DrBob222*

**Math**

27g / 4g/cal = 6.75 Cal. of carbs. 121 - 6.75 = 114.25 Cal. from other sources.
*Saturday, December 29, 2012 at 3:15pm by Henry*

**physics**

There will be a heat release of 540 cal/gm due to condensation and another 65.7 calories/gram due to cooling. 26.4 g x (605.7 cal/g) = ___ cal
*Sunday, November 29, 2009 at 12:56am by drwls*

**Physics - calorie**

The trick is in the definition of specific heat, which is cal/(g-°C). Let's look at the units for : cm(T change) = Q c=cal/(g-°C) m=g T change = °C So the units of Q should be: cal/(g-°C) * g * °C =cal Note: g and °C cancel out to leave calories.
*Sunday, September 20, 2009 at 4:45pm by MathMate*

**Physics**

How much heat is required to melt 50 g of mercury at -45 degrees celsius to 425 degrees celsius. (melting point=-38.degree-c, specific heat=0.03325 cal/g-degree c for solid mercury, .2988 cal/g-c degree c for liquid mercury and .2486 cal/g-c degree for gaseous mercury, heat of...
*Monday, February 27, 2012 at 6:32pm by Abby*

**Physics **

How much heat is required to melt 50 g of mercury at -45 degrees celsius to 425 degrees celsius. (melting point=-38.degree-c, specific heat=0.03325 cal/g-degree c for solid mercury, .2988 cal/g-c degree c for liquid mercury and .2486 cal/g-c degree for gaseous mercury, heat ...
*Monday, February 27, 2012 at 8:06pm by Abby*

**Physics **

How much heat is required to melt 50 g of mercury at -45 degrees celsius to 425 degrees celsius. (melting point=-38.degree-c, specific heat=0.03325 cal/g-degree c for solid mercury, .2988 cal/g-c degree c for liquid mercury and .2486 cal/g-c degree for gaseous mercury, heat ...
*Monday, February 27, 2012 at 8:13pm by Abby*

**chemistry**

How much mass of ice is there? You need the specific heats of ice and of water; the heat of vaporization, and the heat of fusion to answer this question. C,ice = 0.5 Cal/g C C,water = 1.0 Cal/g C H,fusion = 80 Cal/g H,vaporization = 540 Cal/g For Joules, multiply each by 4.184...
*Thursday, March 18, 2010 at 2:25am by drwls*

**Chem**

I wish teachers would get away from Calories vs calories. A Calorie (with a capital C, sometimes called a "big calorie" actually is a kilocalorie and should be addressed as such. So 1 Cal = 1000 calories. There are 4.184 joules in 1 calorie. You can convert from anything to ...
*Sunday, October 16, 2011 at 8:13pm by DrBob222*

**Math**

120 cal/g x 4g = ?? multiply the numbers. 120 x 4 = 480 multiply the units. (cal/g) x g = cal So the answer is 480 cal. I don't know "what the proper units are" unless you mean to evaluate the units at the same time.
*Saturday, January 12, 2008 at 10:16pm by DrBob222*

**physics**

Each gram of 0 deg. C ice requires 80 Cal to melt, 100 Cal to heat from 0 to 100 C as liquid water, and 540 cal to vaporize. That is a total of 720 calories. Multiply that by the number of grams for the answer.
*Friday, February 19, 2010 at 11:03pm by drwls*

**Chemistry**

Yup, as it turns out: cal = calorie Cal = big calorie / kilocalorie No wonder my answers were all wrong - I was converting to calorie instead of kcal! The correct answer was 4.57 Cal.
*Friday, October 30, 2009 at 7:22pm by .*

**chemistry**

Three steps are involved: steam changing to water at 100 C; water at 100 C cooling to 0 C; water at 0 C freezing to ice at 0 C. Steam changing to water: Heat of vaporization of water is 539.4 cal/g Heat = Heat of Vap. x mass = 539.4 cal/g x 75.0 g = 40,500 cal Cooling water ...
*Thursday, January 30, 2014 at 12:22pm by Elina*

**Physics**

A 200g block of copper at 90o C is dropped into 400g of water at 27o C contained in a 300g glass beaker at 27o C. What is the final equilibrium temperature of the mixture? cCu = 0.0924 cal/g°C cglass = 0.2 cal/g°C cwater = 1 cal/g°C
*Saturday, November 30, 2013 at 5:17pm by Angie*

**Chemistry**

If you have the q cal of a reaction, then obviously you can convert to q chem by negating your q cal. But what if you need to get to q rxn (q of the reaction)? Also, my textbook is claiming that the q rxn = -(q sol + q cal). How does that figure into this?
*Thursday, April 1, 2010 at 6:12pm by Fred*

**Chemistry**

What quantity of heat is necessary to convert 50.0 g of ice at 0.0 C into steam 100,0 C? The heat of fusion is 80.0 cal/g, the heat of vaporization is 540 cal/g, and the specific heat of water is 1.00 cal/gC.
*Tuesday, March 27, 2012 at 6:08pm by Jill*

**Chemistry**

Calculate S (measure of total entropy) for the following reaction at 25C and 1atm, and tell whether the entropy is increasing or decreasing. C3H8(g)+5O2(g)-> 3CO2(g)+4H2O(g) S of C3H8=64.5 cal/(mol*K) S of O2= 49 cal/(mol*K) S of CO2= 51.1 cal/(mol*K) S of H2O= 45.1 cal/(...
*Monday, November 16, 2009 at 8:45am by Christina*

**Chemistry Help!!!!**

Calculate S (measure of total entropy) for the following reaction at 25C and 1atm, and tell whether the entropy is increasing or decreasing. C3H8(g)+5O2(g)-> 3CO2(g)+4H2O(g) S of C3H8=64.5 cal/(mol*K) S of O2= 49 cal/(mol*K) S of CO2= 51.1 cal/(mol*K) S of H2O= 45.1 cal/(...
*Monday, November 16, 2009 at 1:43pm by Christina*

**Chemistry**

Calculate S (measure of total entropy) for the following reaction at 25C and 1atm, and tell whether the entropy is increasing or decreasing. C3H8(g)+5O2(g)-> 3CO2(g)+4H2O(g) S of C3H8=64.5 cal/(mol*K) S of O2= 49 cal/(mol*K) S of CO2= 51.1 cal/(mol*K) S of H2O= 45.1 cal/(...
*Monday, November 16, 2009 at 9:03pm by Christina*

**science**

an aluminium container of mass 100g contains 200 g of ice at -20'c heat is added to the system at the rate of 100 cal/sec. what will be the final tmperature of the mixture after 4 min? given:specific heat of ice:0.5 cal/gm'c ,latent heat of fusion:80 cal/gm and specific heat ...
*Thursday, March 17, 2011 at 8:32pm by anu*

**Chem**

How do you convert? (a) 4.50 Cal to calories (b) 600.0 Cal to kilojoules (c) 1.000 J to calories (d) 50.0 Cal to joules
*Sunday, October 16, 2011 at 8:13pm by Miranda*

**Chemistry**

I keep getting the wrong answer :/ I did: 1.28lb x 453.59g/lb= 580.5952g CH4 580.5952g Ch4 x 11.97 cal/g= 6949.724544 cal 6964.724544 cal= 127,000 x 1 cal/g x (Tf-21.5) Final Temp= 21.6*C but this answer was wrong
*Saturday, February 11, 2012 at 12:06am by Heather*

**Basic Math**

Calories/mile is (540 Cal/h)/(5.5 miles/h) = 98.2 Cal/mile The distance run in 3.5 hours is 5.5 mi/h * 3.5 h = 19.8 mi Total Cal = 98.2 Cal/mile * 19.8 mile = __? I am assuming that 3.5 hours is spent running in one week, not per day for a week.
*Friday, December 5, 2008 at 7:47am by drwls*

**Chemistry**

Calculate the heat released when 10.0 g of water at 25.0°C cools to ice at 0.0°C. The specific heat of water is 1.00 cal/(g × °C); the heat of fusion is 80.0 cal/g; and the heat of vaporization is 540.0 cal/g.
*Saturday, April 17, 2010 at 12:35am by helpanderson*

**physics**

how much heat is required to vaporize 7 grams of ice initially at 0 degrees celsius when the latent fusion of ice is 80 cal/g, the vaporization of water is 540 cal/g, and the specific heat of water is 1 cal/(g x C)?
*Friday, February 19, 2010 at 11:03pm by elisabeth*

**Lauren**

memorize the conversion units. 1000J=1kJ 1 cal= 4.1868 J 1k=1000 units so as an example, then on the second 2970kcal(4.1868J/cal) Then immediately check the units, to make certain the units cancel out to leave what you want. In this case... kcal*J/(cal) leaves kJ do the math.
*Thursday, December 5, 2013 at 3:07pm by bobpursley*

**PHYSICS!!**

A Mountain bar has a mass of 6.7×10−2 kg and a calorie rating of 270 cal. What speed would this candy bar have if its kinetic energy were equal to its metabolic energy? [Note: The nutritional calorie, 1 Cal , is equivalent to 1000 calories (1000 Cal ) as defined in ...
*Tuesday, October 16, 2012 at 12:11pm by RoCu*

**physics**

The ice acquires 50 g * 80 Cal/g = 4000 Cal from the water while melting, but remining 0 C. In losing that amount of heat, the original liquid water cools by 4000 Cal/(200g*1 Cal/deg C) = 20 C, leviong it at 20 C. Now you mix 200 g of water at 20 C with 50 g at 0 C. At ...
*Saturday, February 23, 2008 at 9:29am by drwls*

**Biotech**

The dynamic method is used to measure kLa in a fermentor operated at 30degC. Data for dissolved-oxygen concentration as a function of time during the re-oxygenation step is given (table with time (s) and CAL (% air saturation)). The equilibrium concentration of oxygen in the ...
*Tuesday, October 30, 2012 at 6:06pm by MGob*

**Chem**

I'm going to do some assuming here. 2600= 2,600 cal 7.0×10^4=7.0×10^4 g and 20=20ºC Use the following formula: q=mc∆T Where q=2,600 cal m=7.0×10^4 g c= 1.00 cal/g °C ∆T=Tf-Ti=Tf-20ºC solve for Tf q/mc=Tf-20ºC {2,600 cal/[(7.0×10^4 g)*(1.00 cal/g°C)]}+20ºC=Tf *****...
*Friday, March 15, 2013 at 2:07pm by Devron*

**physics**

You want to remove M C (delta T) = 300 g*1.00 Cal/(g C) * 29 C = 8700 calories from the coffee. If m is the mass of ice you add, the heat absorbed by the ice while melting and incresing in temperature from -20 to +58 C is m*Cice*(20) + mCwater*58 + m*80 Cal/g The specific heat...
*Thursday, November 27, 2008 at 12:20pm by drwls*

**Physics 11th grade 'Heat' .**

1) 100g*(80 cal/g + 23 C*1.0cal/gC) = 10,300 cal 2) Additional heat will be transferred to water in the bucket that does not vaporize. You need to know how much water that is. The question is poorly explained. They probably want you to answer 10g*(79C*1.0 cal/gC+540cal/g)
*Friday, March 15, 2013 at 8:40am by drwls*

**chemistry**

A pound of body fat stores an amount of chemical energy equivalent to 3500 Cal. When sleeping, the average adult burns or expends about 0.45 Cal/h for every pound of body weight. How many Calories would a 135 lb person burn during 7 hours of sleep? (Note that 1 Cal = 1 kcal.)
*Saturday, September 11, 2010 at 10:17pm by adrienne*

**science**

A pound of body fat stores an amount of chemical energy equivalent to 3500 Cal. When sleeping, the average adult burns or expends about 0.45 Cal/h for every pound of body weight. How many Calories would a 126 lb person burn during 7 hours of sleep? (Note that 1 Cal = 1 kcal.)
*Thursday, February 10, 2011 at 10:30pm by Renee*

**physics**

A pound of body fat stores an amount of chemical energy equivalent to 3500 Cal. When sleeping, the average adult burns or expends about 0.45 Cal/h for every pound of body weight. How many Calories would a 196 lb person burn during 7 hours of sleep? (Note that 1 Cal = 1 kcal.)
*Thursday, April 28, 2011 at 4:36pm by gabby*

**math**

if a person weighing 150 pounds runs in place for 15 minutes and the reg cal burned is 650 cal /hr Since 15 minutes is 1/4 of an hour, multiply 650 by 1/4. 162.5 cal/hr
*Friday, March 16, 2007 at 8:55pm by patti help*

**Physics**

The work done becomes the same as heat addition, Q = 730J = 174.5 calories. Q = C M *(delta T) C = 1.0 cal/(g*degC) M = 80 g Solve for deltaT, the water temperature change, in degrees C deltaT = 174.5 cal/[(1.0 cal/degC)*(80 g)] = 2.2 deg C
*Wednesday, October 19, 2011 at 12:27am by drwls*

**Chemistry **

I don't know if this is a nutrition course or not. Sometimes we don't know if a Cal is a kcal or not. I will assume the question deals with 1 cal = 4.184 J. q = mass water x specific heat water x delta T. q = 50 x 1 cal/g x (42-18) = 1,200 calories. Then 1,200/0.25 = 4,800 ...
*Saturday, September 18, 2010 at 8:30pm by DrBob222*

**physics**

100 g * 540 cal/g * 4.18 J/cal = 2.26*10^5 Joules
*Sunday, July 22, 2012 at 7:54pm by drwls*

**math**

How many hours of jogging at 5 1/2mi/h would be needed for a 200 pound person to lose 5 pounds and it takes 3500 cal for one pound 740 cal/h for jogging 5 1/2 mi/h Please post your answer, and we'll be glad to critique it. 750x3500 740x3500=2590 No, not right. If it takes ...
*Friday, March 16, 2007 at 10:09pm by patti help*

**Basic Math**

Hi drwls it is not 540 Cal/h it is 740 Cal/h. thanks anyway!
*Friday, December 5, 2008 at 7:47am by A.W.*

**science**

A pound of body fat stores an amount of chemical energy equivalent to 3500 Cal. When sleeping, the average adult burns or expends about 0.45 Cal/h for every pound of body weight. How many Calories would a 170 lb person burn during 10 hours of sleep? (Note that 1 Cal = 1 kcal.)
*Wednesday, October 5, 2011 at 8:51pm by Anonymous*

**su**

0.45 Cal/hr x 24 hr = 10.8 Cal/day. 10.8 Cal/day x ? day = 3500 = ? Hello Rip Van Winkle.
*Friday, October 26, 2012 at 3:54pm by DrBob222*

**Physics **

What mass of ice at 0 oC must you add to 0.5 kg of water at 22 oC to bring the final temperature of the water to 5 oC? (cice = cvapor = 0.5 cliq , cliq = 1 cal/(g oC), hfusion = 80 cal/g, hvaporiztion = 540 cal/g) A. 5.9 B.71 C 100 D106 E1.7
*Wednesday, July 25, 2012 at 11:07pm by Carrie *

**chemistry**

245 Cal = 245 kcal = 245,000 calories 1 calorie = 4.184 J; therefore, 245,000 cal x (4.184 J/cal) = ? J. ?J x (1 kJ/1000) x (1 lb/14.6E3) = ?
*Tuesday, September 11, 2012 at 11:45pm by DrBob222*

**PHYSICAL SCIENCE**

A pound of body fat stores an amount of chemical energy equivalent to 3500 Cal. When sleeping, the average adult burns or expends about 0.45 Cal/h for every pound of body weight. How many Calories would a 179 lb person burn during 10 hours of sleep? (Note that 1 Cal = 1 kcal.)
*Thursday, February 27, 2014 at 3:55pm by ADRIEANNE GREEN*

**Math**

Where is it that tells that cal/g = cal ? And are we talking about calories and grams? What can I say? I am a right brained person :-) Thanks
*Saturday, August 2, 2008 at 2:31pm by Julie*

**Enviromental Science**

How many cal and K cal are required to evaporate one gram of water at 100 degrees Celcius?
*Sunday, September 26, 2010 at 12:54pm by Cindy*

**Math**

x = cal in an apple x + 30 = cal in a pear 10x = 7(x + 30) 10x = 7x + 210 3x = 210 x = 70 70 cal = apple 100 cal = pear
*Monday, January 10, 2011 at 8:25pm by helper*

**Physics**

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*

**chemistry**

If the heat of fusion for ethyl alcohol is 26 cal/g, how many grams must freeze to release 780 cal?
*Sunday, December 5, 2010 at 11:42pm by jacee*

**Chemistry**

take the heat of vaporization 580 cal/g times 602g = cal in human and do the same for life form
*Friday, January 25, 2013 at 10:11am by Argenis*

**Chem**

1 cal = 4.184 J 1000 cal = 1 kcal.
*Saturday, September 10, 2011 at 10:45am by DrBob222*

**physics**

80 cal/g * 100 g = 8000 cal
*Tuesday, July 31, 2012 at 4:21pm by Damon*

**Physics**

(a) 700 J (b) (Internal Energy Increase)/[(Specific Heat)*Mass] = (700J/4.184 J/cal)/(110g*1.00 cal/g*C) = 1.5 C
*Friday, March 30, 2012 at 6:02pm by drwls*

**physics **

How much energy is required to melt 0.1 kg of water (hfusion = 80 cal/g, hvaporiztion = 540 cal/g) at 0oC.
*Tuesday, July 31, 2012 at 4:21pm by Carrie *

**Chemistry**

heat fusion is 75 cal/g (solid to liquid) heat vaporization is 20 cal/g (liq to vapor) heat sublimation = 95 cal/g (75+20)(solid to vapor) heat deposition, the reverse of sublimation, is -95 cal/g.
*Monday, January 23, 2012 at 8:32pm by DrBob222*

**Math**

chart ingredients proteins------------4 gram per cal carb----------------4 gram per cal fat------------------9 gram per cal A banana has 27 grams of carb. it has a total of 121 calories. How many of its calories come from sources other than carbs?
*Saturday, December 29, 2012 at 3:15pm by Ryan*

**physics**

Q (heat) = M*C*(delta T) = 100,000 g * 1.00 cal/g C * 15 deg = ? cal To get the answer in Joules, multiply the number of calories by 4.18
*Thursday, January 28, 2010 at 8:29pm by drwls*

**chemistry**

26 cal/g x ??g = 780 cal
*Sunday, December 5, 2010 at 11:42pm by DrBob222*

**chemistry**

22 g x 540 cal/g = ? cal
*Tuesday, June 18, 2013 at 3:40pm by DrBob222*

**Social Studies**

http://www.hf.rim.or.jp/~kaji/cal/cal.cgi?1777
*Monday, May 19, 2008 at 5:17pm by Writeacher*

**Chemistry!!**

Yes, you use mcdeltaT. I wouldn't go through the 4.184 step since we know that the specific heat of water is 1 cal/g*C or 4.184 J/g*C. This will give you calories and you divide by 1000 to obtain kcal BUT you call it BIG calories (or just calories or Calories---with a capital ...
*Sunday, May 4, 2008 at 11:43am by DrBob222*

**CHEMISTRY**

Assuming Cal = 4 for protein, 4 for carbohydrates and 9 for fat, then 110 Cal-(7*9)-(9*4)-(4x) And your question would make more sense without the {\rm g} mishmash.
*Saturday, February 1, 2014 at 8:53pm by DrBob222*

**Physcial Science**

Hey There, I have 4 questions that I am totally stumped on... Can anyone help? A 200-gram chunk of stuff rises in temperature by 3oC when you input 1200 calories of heat. What is its specific heat capacity? Answers A)50 cal/goC. B) 2 cal/goC. C) 5 cal/goC. D) 0.02 cal/goC. You...
*Friday, September 23, 2011 at 11:11am by Anonymous*

**Physcial Science**

Hey There, I have 4 questions that I am totally stumped on... Can anyone help? A 200-gram chunk of stuff rises in temperature by 3oC when you input 1200 calories of heat. What is its specific heat capacity? Answers A)50 cal/goC. B) 2 cal/goC. C) 5 cal/goC. D) 0.02 cal/goC. You...
*Friday, September 23, 2011 at 11:20am by Anonymous*

**Physcial Science**

Hey There, I have 4 questions that I am totally stumped on... Can anyone help? A 200-gram chunk of stuff rises in temperature by 3oC when you input 1200 calories of heat. What is its specific heat capacity? Answers A)50 cal/goC. B) 2 cal/goC. C) 5 cal/goC. D) 0.02 cal/goC. You...
*Friday, September 23, 2011 at 11:37am by Anonymous*

**physics**

Look up the specific heats of ice, water and steam. Call them Ci, Cw and Cs. Cw is 1.00 cal/g C You will also need the heat of fusion Hf = 80 cal/g and the heat of vaporization Hv = 540 cal/g Q required = 5g * [15Ci + 100Cw + 50 Cs + Hf + Hv] calories
*Tuesday, February 7, 2012 at 8:18am by drwls*

**Algebra 1**

X Scientific cal. Y Graphic cal. Eq1: X + Y = 41. Eq2: 9X + 55Y = 1519. Multiply both sides of Eq1 by -9: -9X - -9Y = -369, 9X + 55Y = 1519 Add the Eqs: 46Y = 1150, Y = 1150 / 46 = 25 Graphic cal. X + 25 = 41, X = 41 -25 = 16 Scientific cal.
*Friday, February 4, 2011 at 12:34am by Henry*

**Physics - ice to liquid**

If you want to work it with calories, the specific heat of water is 1.0 calorie/gram and the heat of fusion for ice is 80 cal/g, therefore mass H2O x 1 cal/g x 80 = 4,000 calories. 4000 calories/80 cal/g = 50 g ice.
*Thursday, September 24, 2009 at 5:48pm by DrBob222*

**Chemistry**

1.28 lbs x 453.59 g/lb = ?grams CH4. ?grams CH4 x (11.97 cal/g) = ? cal heat produced by the l.28 lbs CH4. ?cal = mass H2O x specific heat H2O x (Tfinal-Tintial) Substitute 127,000 for mass H2O Solve for Tf Substitute 21.5 for Ti Substitute 1 cal/g for specific heat H2O.
*Saturday, February 11, 2012 at 12:06am by DrBob222*

**chem/ health**

if it takes alcohol 185 cal to evaporate and water 552 cal to evaporate, which is better to use to reduce a fever?
*Monday, September 22, 2008 at 5:31pm by heather*

**chemistry**

q = mass x specific heat water x delta T q = 25.0g x 1 cal/g*C x (25.7-12.5) = ? For joules, substitute 4.18 J/g*C for 1 cal/g*C
*Sunday, January 22, 2012 at 6:58pm by DrBob222*

**Pre-Cal**

Yes I am, but I don't understand it. I don't have a teacher to teach me. I do my work online, and I just need 2nd semester of pre-cal to graduate.
*Saturday, March 17, 2012 at 9:29pm by Adrianna*

**Chemistry**

convert 175f to c convert 325f to k covnert 575 cal to J canvert 23 Cal to J
*Monday, February 18, 2008 at 11:32am by lizzie*

**Chemistry**

1000 J = 1 kJ is the factor so 220 J x (1 kJ/1000 J) = ? You see you multiply the value you start with by the factor. You can place the factor as 1000/1 or as 1/1000. Only one way is right. The right way will ALWAYS cancel the unit you don't want and keep the unit you want. ...
*Tuesday, March 18, 2014 at 11:02pm by DrBob222*

**physical science **

A sample of silver at 20.0 °C is heated to 100 °C when 1000 cal is added. What is the mass of the silver? (csilver = 0.056 cal/deg-gm)
*Tuesday, March 16, 2010 at 10:13pm by Anonymous*

**physics**

2000 g *[(1.0 cal/C g)*15C +80 cal/g] = 2000*95 = 190,000 calories = 1.65*10^5 Joules = 46.0 watt-hours = 0.046 kWh
*Wednesday, March 9, 2011 at 3:04am by drwls*

**Chemistry**

29,500 J/K x 2.75 K = about 81,000 J(approximate) from the combustion. 81,000 J/4.90 = about 16,000 J/g. 16,000 J/g x (1 cal/4.184 J) = about 4,000 cal. 4,000 cal = 4 kcal = about 4 nutritional calories.
*Friday, February 10, 2012 at 3:10am by DrBob222*

**pre cal**

eh? plug in 6 for x. What do you get? If you're taking pre-cal, this should be a no-brainer.
*Wednesday, June 19, 2013 at 3:19pm by Steve*

**Physics**

Please help! Suppose that a 400 gram copper object (specific heat 0.092 cal/g/degrees Celsius) receives 900 cal. of heat. What is the temperature change? Explain your reasoning. Thanks
*Wednesday, March 2, 2011 at 12:54am by A*

**CHEMISTRY**

The grams of protein in 1 cup of soup that has 110 Cal{\rm Cal} with 7 g{\rm g} of fat and 9 g{\rm g} of carbohydrate. Express your answer using one significant figure.
*Saturday, February 1, 2014 at 8:53pm by LISA*

**chemistry**

Mass is measured in grams, not cal/gram celsius. Specific heat is measured in cal/g*C. . q = mass x specific heat x delta T. Look up specific heat for N2 gas, substitute 89 for delta T and 350 cal for q1, calculate mass in grams.
*Thursday, August 30, 2012 at 12:07am by DrBob222*

**physics(heat)**

Let the final temperature of the mixture = 100ºC. Heat gained by ice and iced water Q1 is Q1 = m•λ + m•c•ΔT = 200•80 + 200•1•100 = 3600 cal. Q2 = (m1•c +W) •(100º -0º) =(250•1 + 50) •100 = 30000 cal. Q =Q1+Q2 = 36000 +30000 = 66000 cal. If entire steam...
*Tuesday, May 29, 2012 at 1:03am by Elena*

**Science**

You've done the hard part; i.e., the calculation. I checked the first one and it's ok. I didn't check the other two. Look at the numbers. methyl alcohol = 1481 cal/g ethyl alcohol = 1848 cal/g propyl alcohol = 2623 cal/g Per gram of alcohol, which produces more heat? Look at ...
*Monday, November 8, 2010 at 3:50pm by DrBob222*

**Physics**

Please help! Suppose that a 400 gram copper object (specific heat 0.092 cal/g/degrees Celsius) receives 900 cal. of heat. What is the temperature change? Explain your reasoning. Do not use algebra Thanks!
*Tuesday, March 1, 2011 at 8:25pm by A*

**Physics**

Q=λ•m 1 J =0.2388 Cal Heat of vaporization for water λ = 2257•10^3•0.2388 =5.39•10^5 Cal/kg, Q=λ•m =5.39•10^5•25 =1.35•10^7 Cal = = 1.35•10^4 kCal
*Monday, April 2, 2012 at 12:22pm by Elena*

**physics**

Q=λm=334 000 J/kg •100 kg =33400000 J=3.34•10^7J=7977452.9 cal= 7.98 •10^6 cal
*Sunday, June 17, 2012 at 9:53am by Elena*

Pages: **1**