Entropy integration problem

[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

 

[wais]

To solve this problem, we can use the formula for the change in entropy, ΔS = Q/T, where Q is the energy absorbed as heat and T is the temperature in Kelvin. We already know the value for Q from part a) of the problem, which is 57900 J. Now, we need to find the temperature in Kelvin at both the initial and final states.

Initial temperature, T1 = 25 + 273 = 298 K
Final temperature, T2 = 100 + 273 = 373 K

Substituting these values into the formula, we get:

ΔS = 57900 J / (373 K) - 57900 J / (298 K) = 155.09 J/K

Therefore, the change in entropy for the 2 kg block of copper is 155.09 J/K. This means that the disorder or randomness of the system has increased by this amount as the temperature was increased from 25°C to 100°C.