A block of ice of mass 120 g at temperature 0 ° C is put in 300 g of water at 25 ° C. The x g of ice melts as the temperature of the water reaches 0 ° C. The value of x is _____________. [Use specific heat capacity of water = 4200 Jkg ⁻¹K⁻ 1, Latent heat of ice = 3.5 × 105 Jkg ⁻ 1] Numerical answer type. Work it out, then open the answer.

Answer: 90. Heat given by water cooling to 0°C melts the ice: mw c Δ T = xL 0.3 × 4200 × 25 = 31500 J x = 315003.5 × 10⁵ = 0.09 kg (less than 120 g, so valid) So x = 90 g.

A block of ice of mass 120 g at temperature 0 ° C is put in 300 g of… | JEE Main 2022 (25 Jul, Shift 2)

Q26JEE Main 2022 · 25 JUL · SHIFT 2Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat

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 $^{-1}\text{K}^{-}$ 1, Latent heat of ice = 3.5 $\times$ 105 Jkg $^{-}$ 1]

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 90

Heat given by water cooling to $0^{\circ}\text{C}$ melts the ice: $m_w c\,\Delta T = xL$
$0.3 \times 4200 \times 25 = 31500\ \text{J}$
$x = \dfrac{31500}{3.5 \times 10^{5}} = 0.09\ \text{kg}$ (less than 120 g, so valid)
So $x = 90\ \text{g}$ .

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