Calorimetry

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Question

What will be the temperature, when 150 g of ice at 0°C is mixed with 300 g of water at 50°C? Specific heat of water =1 ca lg-1°C. Latent heat of fusion of ice = 80 -1 ca lg^-1.

Moderate

Solution

Let us assume that T > 0° C
Heat lost by water = Heat gained by ice to melt + heat gained by water formed from ice
$\begin{array}{l} 300 \times 1 \times (50 - T) = 150 \times 80 + 150 \times 1 \times (T - 0) \\ \Rightarrow T = 6 {.7}^{\circ} C \end{array}$
Hence, our assumption that T > 0°C is correct.

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