JEE Main 2019 Physics question
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/°C, Specific heat of Ice = 2.1 J/g/°C, Heat of fusion of water at 0°C = 334 J/g)
Answer: (D)
Solution
50 × 4.2 × 40 = 8400 J = m(2.1)(20) + (m − 20)(334) ⇒ m ≈ 40 g.