- #1

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## Homework Statement

The specific heat capacity of a metal at low temperature (T) is given as ##C_p = 32\left({\dfrac{T}{400}}\right)^3## (kJK

^{-1}kg

^{-1}). A 100g vessel of this metal is to be cooled from 20 K to 4 K by a special refrigerator operating at room temperature (27°C). The amount of work required to cool the vessel is

(A) equal to 0.002 kJ

(B) greater than 0.148 kJ

(C) between 0.148 kJ and 0.028 kJ

(D) less than 0.028 kJ

## Homework Equations

q = u + w

dq = mC

_{p}dT

## The Attempt at a Solution

I don't know how to calculate the change in internal energy. I do know how to calculate q.

-dq = mC

_{p}dT

##q = -\int\limits_{20}^{4}32\left({\dfrac{T}{400}}\right)^3 dT##

q = 0.002 kJ

(A) is not the answer (and it shouldn't be because I've ignored Δu). How do I find Δu so I can then find the work done from the first law equation? Please help.