Entropy

[Quipzley]

Find a) the energy absorbed as heat and b) the change in entropy of a 2 kg block of copper whose temperature is increased reversibly from 25 degrees C to 100 degrees C. The specefic heat of copper is 386 J/kg * K.

a) Q = 386 J/kg *K * (373 K - 298 K) * 2 kg = 57900 J

b) the change in entropy = integral(dQ / T). I'm unsure of how to integrate to be able to have a usable formula???

[Andrew Mason]

Find a) the energy absorbed as heat and b) the change in entropy of a 2 kg block of copper whose temperature is increased reversibly from 25 degrees C to 100 degrees C. The specefic heat of copper is 386 J/kg * K.

b) the change in entropy = integral(dQ / T). I'm unsure of how to integrate to be able to have a usable formula???

Since \Delta Q = cm\Delta T:

\Delta S = \int_{Ti}^{Tf }\frac{\delta Q}{T} = \int_{Ti}^{Tf } \frac{cm\delta T}{T}

Note:
\int_{Ti}^{Tf }\frac{\delta T}{T} = ln(\frac{T_f}{T_i})

AM