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toothpaste666
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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?