Question

0.100 kg of water at 10∘C is added to 0.300 kg of soup at 50∘C. Assume complete transfer of thermal energy from soup to the water, with no transfer of energy to the environment. Specific heat of water is 4180 J/kg⋅∘C. The soup has the same specific heat as water.

A) Determine the final temperature? Express in Celsius

B) Estimate the entropy change of this water-soup system during the process using the actual temperatures to determine the heat transferred and the average temperatures to determine the entropy change? Express in dimension of entropy (J/K)

Answer #1

**let m1 = 0.1 kg
T1 = 10 C = 10 + 273 = 283 K
m2 = 0.3 kg
T2 = 50 C = 50 + 273 = 323 K
C = 4180 J/(kg C)
A)
let T3 is the final temperature.**

**heat gained by water = heat lost by soup**

**m1*C*(T3 - T1) = m2*C*(T2 - T3)**

**0.1*(T3 - 10) = 0.3*(50 - T3)**

**==> T3 = 40 degrees celcius
<<<<<<<------Answer**

**B) T3 = 40 + 273**

**= 313 K**

**Total change in entropy = m1*C*ln(T3/T1) +
m2*C*ln(T3/T2)**

**= 0.1*4180*ln(313/283) +
0.3*4180*ln(313/323)**

**= 2.68 J/k
<<<<<<<------Answer**

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