Two blocks of different temperatures are brought together. Block A is a 360-kg b

Question

Two blocks of different temperatures are brought together. Block A is a 360-kg block is made of Material X which has a temperature-dependent specific heat. Its initial temperature is T_a = 200degree C. Block B is a 10^3-kg iron block with an initial temperature T_B = 27degree C. The specific heat of Material X in the range of interest is c_A = T*0.64 J/(kg K^2) and the specific heat of iron is C_B = 452 J/(kg K). We bring the two blocks together and wait for them to reach a common equilibrium temperature T_f. If we assume that the two block system does not lose energy to the environment, then what is the change in entropy of the two block system? 0degreeC = 273.15 K delta S =

Explanation / Answer

Heat lost by material A is gained by B

MA * int ( CA * dt ) = MB * CB * ( Tf - 27)

int ( CA * dt ) from Tf to 200 = 0.64 T^2/2 = 0.32 * Tf^2 - 0.32 * 40000

The equation becomes,

360 * ( -0.32 * Tf^2 + 0.32 * 40000) = 103 * 452 * ( Tf - 27)

Tf =100.82 C

Change in entropy for A = Cp ln ( Tf/ Ti)

dSa = 0.64 * ln(100.82 + 273)/(200+273)

dSa = -0.15

dSb Change in entropy for B = Cp ln ( Tf/ Ti)

dSb = 0.64 * ln(100.82 + 273)/(27+273)

total entropy = 0.15 - 0.14 = - 0.01

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