# Calorimetry problem

1. Oct 31, 2014

### toothpaste666

1. The problem statement, all variables and given/known data
A glass cup with a mass of 0.1 kg and an initial temperature of 23◦C is filled with 0.3 kgof water at 80◦C

A)What is the final temperature of the water and cup?
B)How much heat must be added to raise the temperature to 90◦C?
C) What is the change in the internal energy of the water?
2. Relevant equations
E = Q - W
Q = mcdeltaT

3. The attempt at a solution
A) heat gained by glass = heat lost by water

$m_{glass}c_{glass}(T_f - 23 C) = m_{water}c_{water} (80 C - T_f)$

where cglass = specific heat of glass = 840 J/kgC
cwater = spec heat of water = 4186 J/kgC
and Tf is the final temp

$(.1 kg) (840 J/kgC) (T_f - 23 C) = (.3 kg)(4186 J/kgC)(80 C - T_f)$

$84T_f - 1932 = 100464 - 1256T_f$

$1340T_f = 102396$

$T_f = 76.4 C$

B) Q = heat gained by glass + heat gained by water

$Q = m_{glass}c_{glass}(T_f - 76.4) + m_{water}c_{water}(T_f - 76.4)$

$Q = (T_f - 76.4)(m_{glass}c_{glass} + m_{water}c_{water})$

$Q = (90 - 76.4)(.1(840)+ .3(4186))$

$Q = (13.6)(84 + 1256)$

$Q = 18221 J$

C) since no work is done, the internal energy of the water would be the same as Q for the water
$E = Q = m_{water}c_{water}(80 C - 76.4 C)$

$E = Q = .3(4186)(80 C - 76.4 C)$

$E = Q = .3(4186)(3.6)$

$E = Q = 4520 J$

is this the correct solution?

2. Oct 31, 2014

### ehild

The question is the change of internal energy, ΔE, which is equal to the added heat . It is not the same as the internal energy E.
But ΔE has sign, you have to indicate.

3. Nov 1, 2014

### toothpaste666

the water loses heat to reach the final temp so it would be negative?

4. Nov 1, 2014

Yes.

5. Nov 1, 2014

Thank you!