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

[toothpaste666]

Homework Statement

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?

Homework Equations

E = Q - W
Q = mcdeltaT

The Attempt at a Solution

A) heat gained by glass = heat lost by water

[itex] m_{glass}c_{glass}(T_f - 23 C) = m_{water}c_{water} (80 C - T_f) [/itex]

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

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

[itex] 84T_f - 1932 = 100464 - 1256T_f [/itex]

[itex] 1340T_f = 102396 [/itex]

[itex] T_f = 76.4 C [/itex]

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

[itex] Q = m_{glass}c_{glass}(T_f - 76.4) + m_{water}c_{water}(T_f - 76.4) [/itex]

[itex] Q = (T_f - 76.4)(m_{glass}c_{glass} + m_{water}c_{water})[/itex]

[itex] Q = (90 - 76.4)(.1(840)+ .3(4186))[/itex]

[itex] Q = (13.6)(84 + 1256)[/itex]

[itex] Q = 18221 J [/itex]

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

[itex] E = Q = .3(4186)(80 C - 76.4 C) [/itex]

[itex] E = Q = .3(4186)(3.6) [/itex]

[itex] E = Q = 4520 J[/itex]

is this the correct solution?

 

[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.

 

[toothpaste666]

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

 

[ehild]

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

Yes.

 

[toothpaste666]

Thank you!