# How do I find the specific heat of a hot object in water?

1. Apr 6, 2015

### Eclair_de_XII

1. The problem statement, all variables and given/known data
"Using the above calorimetre with 100.0 g of water of in it at 20.0°C, a 50.00 g sample of a metal at 100.0°C is added. The temperature of the calorimetre rises to 27.2°C. What is the specific heat? (S.H. = 0.20 cal/g°C)"

2. Relevant equations
Heat capacity of calorimetre used = 8.0 cal/°C

3. The attempt at a solution
(27.2 - 20)°C ⋅ 8.0 cal/°C = 57.6 cal
100.0 g ⋅ 7.2°C = 720 calories absorbed by water
50.00 g ⋅ (27.2 - 100) = -3640 calories released by metal
-3640 + 720 = -2920 calories absorbed by calorimetre
(-2920 cal/7.2°C⋅100 g) = 4.05555... cal/g°C ≠ (0.20 cal/g°C)
They already gave me the answer (0.20 cal/g°C) but they want a process, and I am almost sure I am calculating the last step wrong.

2. Apr 6, 2015

### RaulTheUCSCSlug

This is more of a chemistry problem, but you are going to need to use the specific heat capacity formula, do you know any other formulas in your book/notes?

3. Apr 6, 2015

### Eclair_de_XII

Yes, I originally posted it in that forum, but for some reason, it got moved here. In any case, here are most of the basic formulas listed:

Heat capacity (J/°C) = S.H.(J/g°C) x mass (g)
q = S.H.(J/g°C) x mass (g) x Δtemperature (°C)
or
q = m x SH x ΔT
Q (lost by metal) = Q (gained by cold water) + Q (gained by calorimetre)

4. Apr 7, 2015

### SteamKing

Staff Emeritus
No, this is not a chemistry problem at all. There is no discussion of what the metal might be.

If you look at other posts in the Intro Physics HW forum, you'll see plenty of calorimeter problems similar to this one.

The problem which the OP has is that he is claiming the metal loses 3640 calories while it cools, but that figure presupposes that the specific heat of the metal is 1 cal/g-°C, which it clearly is not.

The amount of heat which the metal loses is equal to the amount of heat gained by the water and the calorimeter after the metal is inserted, no more, no less.

5. Apr 7, 2015

### Eclair_de_XII

(1 cal/g°C) ⋅ 100.0 g ⋅ 7.2°C = 720 calories absorbed by water
(x cal/g°C) ⋅ 50.00 g ⋅ (27.2 - 100) = -3640x calories released by metal
(27.2 - 20)°C ⋅ 8.0 cal/°C = 57.6 calories absorbed by calorimetre
720 + 57.6 = -3640x
x = -0.2136 cal/g°C

I'm not sure if this is correct...

6. Apr 7, 2015

### SteamKing

Staff Emeritus
Specific heat capacity is never negative.

It's better to use the formula q = m C ΔT when calculating the specific heat capacity. Here, C is the specific heat capacity.

q = q(water) + q(calorimeter) = m(sample) × C × ΔT

Fill in the known quantities and solve for the unknown value of C.

q is going to be equal to the heat absorbed by the water and the calorimeter. All of this heat is assumed to be transferred from the hot metal sample placed inside the calorimeter.

7. Apr 7, 2015

### Eclair_de_XII

q = 720 cal + 57.6 cal = 50 g × x-cal/g°C × 72.8°C
q = 777.6 cal = 3640x
q = 0.2136 cal
x = 0.2136 cal/g°C

8. Apr 7, 2015

### SteamKing

Staff Emeritus
Looks good. You might also express the specific heat of the same with a precision consistent with the data given in the OP.

9. Apr 7, 2015

### Eclair_de_XII

I do not understand what that would entail.

10. Apr 7, 2015

### SteamKing

Staff Emeritus
Significant figures. Have you been taught how to use them?

11. Apr 7, 2015

### Eclair_de_XII

I'm guessing three sig-figs? In any case, it still doesn't match the answer given.

x = 0.214 cal/g°C

12. Apr 7, 2015

### SteamKing

Staff Emeritus
The heat capacity of the calorimeter itself (8.0 cal / g-°C) is given to only two significant figures, so your answer for the specific heat of the sample can't be more precise than that.

http://en.wikipedia.org/wiki/Significant_figures

13. Apr 7, 2015

### Eclair_de_XII

x = 0.21?

14. Apr 7, 2015

### SteamKing

Staff Emeritus
Close enough to 0.20.

15. Apr 7, 2015

Thanks.