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Ice at -20 °C is added to 50 g of water at 40 °C. When the temperature of the mixture reaches 0 °C, it is found that 20 g of ice is still unmelted. The amount of ice added to the water was close to (specific heat of water =4.2 J g⁻¹ (° C)⁻¹, specific heat of ice =2.1 J g⁻¹ (° C)⁻¹, heat of fusion of water at 0 °C=334 J g⁻¹)

Asked in JEE Main 11th Jan 1st Shift 2019 · Latent heat and change of state

• (1) 60 g
• (2) 50 g
• (3) 40 g
• (4) 100 g

Step-by-step solution

Why the other options are wrong

• (1) comes from no consistent slip lands here; it is a filler above the key. It is checkable that 60 g of ice at -20 °C would need 60×42+40×334=15880 J to reach this state, while the water can only give 8400 J.
• (2) comes from treating the ice as already at 0 °C and skipping the 2.1 J g⁻¹ (° C)⁻¹ warming stage: then (8400)/(334)=25.1 g melts, and adding the 20 g left over gives 45 g, which a student rounds up to this option.
• (4) comes from forgetting the latent heat altogether and spending all 8400 J on warming the ice, m=(8400)/(2.1×20)=200 g, then halving it to stay on the sheet. The other loose route, melting the whole slab instead of m-20, gives (8400)/(376)=22 g.

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