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Preeti Dabral 3 years ago (2983787)
{tex}\mathop {at\,0^\circ C}\limits^{ice} \to \mathop {at\,0^\circ C}\limits^{water} \to \mathop {at\,6^\circ C}\limits^{water} {/tex}
At equilibrium, heat released by 100gm of water = heat absorbed by 100gm of ice i.e.m1c1 (80 - 6) = m2 L + m2c1(6 - 0)[c1, m1, m2 and L are specific heat of water, mass of water, mass of ice and latent heat of fusion of ice respectively.)
{tex}\therefore{/tex}100 × 1 × 74 = 100 L + 100 × 1 × 6
{tex}\Rightarrow{/tex} L = (1 × 74) -6
{tex}\Rightarrow{/tex}L = 68 cal/gm
This is the required value of latent heat of fusion of ice.
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