PRINTER VERSION BACK NEXT Chapter 20, Problem 070 1.4 g block of silver at-121 a

Question

PRINTER VERSION BACK NEXT Chapter 20, Problem 070 1.4 g block of silver at-121 are placed together in an insulated container. (The specific heat of tungsten as 1343/kg K and the specific heat of silver is 236 3/kg.K.) (a system undergo in reaching the equilibrium temperature? ) What is the equilibrium temperature? What entropy changes do (b) the tungsten, (c) the silver, and (d) the tungsten-silver (a) Number (b) Number (c) Number (d) Number Click if you would like to Show Work for this question: Units Units Units LINK TO 11x INK TO SAMPLE PROBLEM LINK TO SAMPLE PROBLE

Explanation / Answer

given m1 = 44.1 g

T1 = 28.8 C

m2 = 21.4 g

T2 = -121 C

C1 = 134 J/kg K

C2 = 236 J/kg K

a. equilibrium temp = T

form energy balance

m1*C1(T1 - T) = m2*C2(T - T2)

44.1*134(28.8 - T) = 21.4*236(T + 121)

hence

33.698463 - 1.1700855 = T + 121

hence

T = -40.233 C

b. change in entropy of tungsten = S1

now, dS1 = mC1*dT/T

S1 = m*C1*ln(T/T1) ( in kelvin)

hence

S1 = -1.53391 J/K

c. change in entropy of silver = S2 = m2*C2*ln(T/T2) = 2.1504224 J/K

d. total entropy change = S2 + S1 = 0.6165 J/K

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