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April 16, 2014

Search: math/ cal

Number of results: 215,800

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

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

chemistry
How many calories are required to heat 25.0 g of platinum from 24.5 C to 75C (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

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

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

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

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

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

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
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

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 100C is cooled to 10 g of liquid water at 100C? 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

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

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

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.

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

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

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

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
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

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

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

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 .

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
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

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

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

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/gC cglass = 0.2 cal/gC cwater = 1 cal/gC
Saturday, November 30, 2013 at 5:17pm by Angie

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

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

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

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

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

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

Chemistry
Calculate the heat released when 10.0 g of water at 25.0C cools to ice at 0.0C. 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

PHYSICS!!
A Mountain bar has a mass of 6.710−2 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

Chem
I'm going to do some assuming here. 2600= 2,600 cal 7.010^4=7.010^4 g and 20=20C Use the following formula: q=mc∆T Where q=2,600 cal m=7.010^4 g c= 1.00 cal/g C ∆T=Tf-Ti=Tf-20C solve for Tf q/mc=Tf-20C {2,600 cal/[(7.010^4 g)*(1.00 cal/gC)]}+20C=Tf *****...
Friday, March 15, 2013 at 2:07pm by Devron

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

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

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
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

cal Reiny!!!!!!!!!!
I have these answer choices: x=9 and x=6 x=-9 x=-9 and x=6 x= 6 The graph has no horizontal tangents.
Wednesday, October 23, 2013 at 3:15pm by cal

cal
Find dy/dx by implicit differentiation. x^3+8x+x^13y-y^6=4
Thursday, October 24, 2013 at 6:27pm by cal

cal
dy/dx = 3x^2+8+13x^12y / 6y^5-x^13
Thursday, October 24, 2013 at 6:27pm by cal

cal
from the ans choices i choose: dy/dx= - 4x^2/x^2+1 is that right
Thursday, October 24, 2013 at 10:41pm by cal

cal
Use the Quotient Rule to differentiate the function. f(x)= 8x/x^5+3
Tuesday, October 22, 2013 at 10:51pm by cal

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

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

cal
Determine all values of x, (if any), at which the graph of the function has a horizontal tangent. y(x) = 6x/(x-9)^2
Wednesday, October 23, 2013 at 3:15pm by cal

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

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

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

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

cal
Find the slope of the graph of the function at x = 5. f(x)= -2x^2+6/x^2
Tuesday, October 22, 2013 at 9:09pm by cal

cal
Use the Product Rule to differentiate f(u)= √u (5-u^6)
Thursday, October 24, 2013 at 12:03pm by cal

cal
Find the derivative of the function. f(x)= x^8√5-3x
Thursday, October 24, 2013 at 2:23pm by cal

cal
but the ans is f(x)= x^7(80-51x)/2√5-3x
Thursday, October 24, 2013 at 2:23pm by cal

cal
Find the second derivative of the function. f(x)= (3x^3+7)^7
Thursday, October 24, 2013 at 4:48pm by cal

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

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

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

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

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

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

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

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

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

cal
I got it. Never mind.
Tuesday, October 22, 2013 at 10:51pm by cal

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

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

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

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

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

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

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

Math.
420cal = 2.25 cans ? cal = 1can To find the number of calories we just do the cross product, so: 420cal x 1can / 2.25cans = ? cal 420cal/2.25cans = 186.6667cal
Thursday, October 28, 2010 at 4:55am by Ivan

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

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

cal
Find dy/dx by implicit differentiation given that tan(4x+y)=4x
Thursday, October 24, 2013 at 10:41pm by cal

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

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

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