What is the change in the internal energy of the water?

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!