# I need help with H=mcDeltaT

1. May 27, 2010

### Kyyliieee

I need help with H=mcDeltaT!!

1. The problem statement, all variables and given/known data

Ok, so im not actually sure if this is technically Physics or not, but I really need help with it!!
I understand that;
H= heat
m=mass
c= heat capacity?
Delta= change
t= temperature
(DeltaT= change in temperature)

What im not so sure on is how exacty to solve it.
It thought I was doing it right but the answers all keep coming out wrong..

2. Relevant equations

I was trying to figure this question out..

Suppose you drink 6 glasses of ice water (each 250g & 0 degrees C) Using H=mcDeltaT
a. How many calories of energy would you burn to bring the up to normal body temperature of 37 degrees C?
b. How many Calories is this?
c. A serving of French fries has 240 Calories. How many glasses of ice water would you have to drink to burn these off? (Use part b to help)

3. The attempt at a solution

I feel completely lost on all of this. My friend in my class said that Part c is 26 glasses of water, but I keep getting absolutly no where near that. So either shes wrong or im worse off then I thought.
Can anybody try to help explain this to me??

2. May 27, 2010

### collinsmark

Re: I need help with H=mcDeltaT!!

Wow. This thread is making me crave McDonalds.

Seriously though, let us know what you go got, and show us how you got it. A couple of things to keep in mind:

calorie (lower case 'c', sometimes called "gram calorie") and
Calorie (upper case 'c', sometimes called "kilogram calorie")
don't mean the same thing. One is 1000 times bigger than the other.

As long as we're on the topic of the letter 'c', the 'c' in your equation $H = mc \Delta T$ is the specific heat (not calories nor Calories. H is measured in calories or Calories). What value are you using for 'c', the specific heat? (Hint: remember, you're probably not trying to find the energy in units of Joules, you probably want to find the energy in units of calories [or Calories], so it is important that 'c' provides the correct conversion. If you're still not sure ask yourself this: "how many calories does it take raise 1 g of water 1o C?" Plug that into your H = mcDeltaT and make sure you're using the right 'c' such that it makes sense.]

3. May 27, 2010

### Kyyliieee

Re: I need help with H=mcDeltaT!!

Well what I did was just H=(1500g)(4.187kJ)(37)=232378.5c or 323.3785C

How I got that was:
1. It said 6 glasses of ice water each one 250g and 0degrees C So I figured if its 6 glasses I need to multiply 250 by 6 equaling 1500

2. Then 4.187kJ cuse in the few notes I do have it says that the specific heat of water is 4.187kJ, but do I need to convert that into calories? (and if I did would that make it 1c?)

3. I got 37 for DeltaT cuse (I think) DeltaT means the change in temperature.
So T2-T1=DeltaT and since temperature2 is 37 and temperature1 is 0, it would just be 37 degrees C

and thats what I did to get my answer, but I know that doesnt look right at all..

4. May 27, 2010

### Staff: Mentor

Re: I need help with H=mcDeltaT!!

Yes, you need to convert between joules and calories. You have ignored units, no wonder your result is off.

1 cal = 4.187 J (beware - 1C = 1000c, 1 kJ = 1000 J)

Note that calorie was defined as amount of heat necessary to heat up 1 g of water by 1 deg C, that makes it a convenient unit when dealing with water.

5. May 27, 2010

### Kyyliieee

Re: I need help with H=mcDeltaT!!

So if I just change 4.187 J to 1c then the equation will be,
H=(1500g)(1c)(37degreesC)= 55,500c or 55.5C
Is that right?

So then the answers will be
a. 55,500c
b. 55.5C
c. 4.32 glasses of water

Cuse for Part c would you just take 240C/55.5C= 4.32 glasses of water??

6. May 27, 2010

### collinsmark

Re: I need help with H=mcDeltaT!!

Yes, except for the units. The specific heat of water, in terms of calories, is

$$c = 1 \left[ \frac{\mathrm{calorie}}{\mathrm{g} \ ^{\circ} \mathrm{C} } \right]$$

That comes from the original definition of calorie. One calorie is the amount of heat it takes to raise one gram of water, one degree C.

So we have

$$H = \left( 1500 \ [\mathrm{g}] \right) \left( 1 \ \left[ \frac{\mathrm{calorie}}{\mathrm{g} \ ^{\circ} \mathrm{C} } \right] \right) \left( 37 \ [^{\circ} \mathrm{C}] \right)$$

$$= (1500)(37) \ [\mathrm{g}] \left[ \frac{\mathrm{calorie}}{\mathrm{g} \ ^{\circ} \mathrm{C} } \right] [^{\circ} \mathrm{C}]$$

$$= 55500 \ [\mathrm{g}] \left[ \frac{\mathrm{calorie}}{\mathrm{g} \ ^{\circ} \mathrm{C} } \right] [^{\circ} \mathrm{C}]$$

$$= 55500 \ \mathrm{calories}$$

Do you see how the units work?

The above is also equal to

$$H = \left( 55500 \ [\mathrm{calories}] \right) \left( \frac{1}{1000} \ \left[ \frac{\mathrm{Calorie}}{\mathrm{calorie}} \right] \right)$$

$$= 55.5 \ \mathrm{Calories}$$

Not quite. What you have just done is calculated how many sets-of-6 glasses of water it takes. So you need to multiply by 6 to get the total number of glasses.

$$glasses = \left( sets \ [\mathrm{set}]} \right) \ \left(6 \ \left[\frac{\mathrm{glass}}{\mathrm{set}} \right] \right)$$

Alternately you could use your [itex] H = mc \Delta T [/tex] equation and solve for m. Then use the equation,

$$glasses = \frac{m \ [\mathrm{g}]}{250 \ \left[ \frac{\mathrm{g}}{\mathrm{glass}}\right] }$$

My point of giving you the above equations is to demonstrate how the units work. Do you see how paying attention to units can help with problems like this? Units are your friends. You can't (and shouldn't) derive equations by analyzing units alone, but the units can (and often do) easily let you know if you messed up with the math somewhere.