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

  • #1
832
45

Homework Statement


"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)"

Homework Equations


Heat capacity of calorimetre used = 8.0 cal/°C

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.
 

Answers and Replies

  • #2
RaulTheUCSCSlug
Gold Member
179
25
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
832
45
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?
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
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,796
1,668
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?
No, this is not a chemistry problem at all. There is no discussion of what the metal might be. :rolleyes:

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

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. :wink:

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. :wink:
 
  • #5
832
45
(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
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,796
1,668
(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...
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
832
45
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
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,796
1,668
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
Looks good. You might also express the specific heat of the same with a precision consistent with the data given in the OP.
 
  • #9
832
45
You might also express the specific heat of the same with a precision consistent with the data given in the OP.
I do not understand what that would entail.
 
  • #10
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,796
1,668
I do not understand what that would entail.
Significant figures. Have you been taught how to use them?
 
  • #11
832
45
I'm guessing three sig-figs? In any case, it still doesn't match the answer given.

x = 0.214 cal/g°C
 
  • #12
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,796
1,668
I'm guessing three sig-figs? In any case, it still doesn't match the answer given.

x = 0.214 cal/g°C
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
 
  • #14
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,796
1,668
  • Like
Likes RaulTheUCSCSlug
  • #15
832
45
Thanks.
 

Related Threads on How do I find the specific heat of a hot object in water?

Replies
3
Views
1K
Replies
3
Views
4K
Replies
9
Views
2K
Replies
8
Views
4K
  • Last Post
Replies
1
Views
1K
Replies
1
Views
474
Replies
1
Views
1K
Replies
3
Views
33K
Replies
6
Views
2K
Top