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 \(grams\))
- A \(61.7\)
- B \(63.8\)
- C \(69.3\)
- D \(64.6\)
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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