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## Homework Statement

A 3.00-kg block of silicon at 60.0°C is immersed in 6.00 kg of mercury at 20.0°C. What is the entropy increase of this system as it moves to equilibrium? The specific heat of silicon is 0.17 cal/(g·K) and the specific heat of mercury is 0.033 cal/(g·K).

## Homework Equations

Q = mCΔT

ΔS = ΔS

_{H}+ ΔS

_{C}

ΔS = Q/T

## The Attempt at a Solution

m

_{Si}= 3.00 Kg

m

_{Hg}= 6.00 Kg

T

_{Si}= 60.0 °C = 333.15 K

T

_{Hg}= 20.0 °C = 293.15 K

C

_{Si}= 0.17 cal/(g⋅K) = 711.62 J/(Kg⋅K)

C

_{Hg}= 0.033 cal/(g⋅K) = 138.138 J/(Kg⋅K)

ΔT = T

_{H}- T

_{C}= 333.15 K - 293.15 K = 40 K

**Q = mCΔT**

Q

_{Si}= (3.00 Kg)(711.62 J/(Kg⋅K))(40 K)

= 85394.4 J

Q

_{Hg}= (6.00 Kg)(138.138 J/(Kg⋅K))(40 K)

= 33153.12 J

**ΔS = Q/T**

(Heat leaves the silicon block, so its change in entropy is negative)

ΔS

_{Si}= -(85394.4 J / 333.15 K)

= -256.324 J/K

(Heat enters the Mercury, so its change in entropy is positive)

ΔS

_{Hg}= 33153.12 J / 293.15 K

= 113.093 J/K

**ΔS = ΔS**

_{H}+ ΔS_{C}ΔS = -256.324 J/K + 113.093 J/K

= -143.231

This is a multiple choice question and my answer does not match any of the five possible choices. However, the question also states that we can use our own answer if we explain why we believe it to be correct, but I do not feel confident that my answer is correct.