The specific heat of water
= 4200 J kg-1K-1 and the latent heat of
ice = 3.4 $\times$ 105 J kg–1. 100 grams of ice at
0oC is placed in 200 g of water at 25oC. The
amount of ice that will melt as the temperature
of water reaches 0oC is close to (in grams) :
Solution
Heat loss by water<br><br>$Q = {m_w}s\Delta \theta$<br><br>$= \left( {{{200} \over {1000}}} \right).(4200)(25) = 21000\,J$
<br><br>This heat will absorbed by the ice and let mass $\Delta$m<sub>i</sub> got melted.
<br><br>So $\Delta {m_i}L = 21000$<br><br>$$\Delta {m_i} = {{21000} \over {3.4 \times {{10}^5}}} \times {10^3}\,gm = 61.7\,grams$$
About this question
Subject: Physics · Chapter: Thermodynamics · Topic: Zeroth and First Law
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