Calculate the change in entropy when 1 g of ice at 0- C is heated to form water at 40- C-

In this case heat is required to melt ice and to rise the temperature to 40^∘ C =mLT+ mC∫ _T_1^T_2d T/T =1× 80273+1× 1× 2.303 log313273 =0.28+0.1366=0.42 cal ^∘ ...

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Standard XII

Physics

2nd Law of Thermodynamics

Calculate the...

Question

Calculate the change in entropy when $1 g$ of ice at $0^{\circ}$ C is heated to form water at $40^{\circ} C$ .

0.28 cal /^{\circ} C

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1.411 cal /^{\circ} C

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0.41 cal /^{\circ} C

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none of these

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Solution

The correct option is C $0.41 cal /^{\circ} C$
In this case heat is required to melt ice and to rise the temperature to $40^{\circ} C$
$= \frac{m L}{T} + m C \int_{T_{1}}^{T_{2}} \frac{d T}{T}$
$= \frac{1 \times 80}{273} + 1 \times 1 \times 2.303 log \frac{313}{273}$
$= 0.28 + 0.1366 = 0.42 cal^{\circ} C^{- 1}$

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Calculate the change in entropy when 1 g of ice at 0∘ C is heated to form water at 40∘ C.

The correct option is C 0.41 cal/∘ CIn this case heat is required to melt ice and to rise the temperature to 40∘C =mLT+mC∫T2T1dTT =1×80273+1×1×2.303 log313273 =0.28+0.1366=0.42 cal ∘C−1

Calculate the change in entropy when $1 g$ of ice at $0^{\circ}$ C is heated to form water at $40^{\circ} C$ .