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JEE Mains · Physics · STD 11 - 10.1, thermonetry,thermal expansion and calorimetry

The specific heat of water \(=4200\, J\, kg ^{-1}\, K ^{-1}\) and the latent heat of ice \(=3.4 \times 10^{5}\, J\, kg ^{-1}.\) \(100\) grams of ice at \(0^{\circ} C\) is placed in \(200\, g\) of water at \(25^{\circ} C\). The amount of ice that will melt as the temperature of water reaches \(0^{\circ} C\) is close to (in \(grams\)) 

  1. A \(61.7\)
  2. B \(63.8\)
  3. C \(69.3\)
  4. D \(64.6\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(61.7\)

Step-by-step Solution

Detailed explanation

Here the water will provide heat for ice to melt therefore \(m_{w} s_{w} \Delta \theta=m_{i c e} L_{i c e}\) \(m _{ ice }=\frac{0.2 \times 4200 \times 25}{3.4 \times 10^{5}}\) \(=0.0617 \,kg\) \(=61.7\, gm\) Remaining ice will remain un-melted
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