A block of ice of mass 120 g at temperature 0$^\circ$C is put in 300 g of water at 25$^\circ$C. The x g of ice melts as the temperature of the water reaches 0$^\circ$C. The value of x is _____________.
[Use specific heat capacity of water = 4200 Jkg$-$1K$-$1, Latent heat of ice = 3.5 $\times$ 105 Jkg$-$1]
Answer (integer)
90
Solution
<p>Heat lost by water = Heat gained by ice</p>
<p>$0.3 \times 4200 \times 25 = x \times 3.5 \times {10^5}$</p>
<p>$x = {{0.3 \times 4200 \times 25} \over {3.5 \times {{10}^5}}}$</p>
<p>$= 90 \times 100 \times {10^5} \times {10^3}$ gram = 90 gm</p>
About this question
Subject: Physics · Chapter: Thermodynamics · Topic: Zeroth and First Law
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