JEE MAIN - Physics (2022 - 25th July Evening Shift - No. 20)
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
90
Explanation
Heat lost by water = Heat gained by ice
$$0.3 \times 4200 \times 25 = x \times 3.5 \times {10^5}$$
$$x = {{0.3 \times 4200 \times 25} \over {3.5 \times {{10}^5}}}$$
$$ = 90 \times 100 \times {10^5} \times {10^3}$$ gram = 90 gm
Comments (0)
