JEE PYQ: Thermodynamics Question 41
Question 41 - 2019 (11 Jan Shift 1)
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)
(1) 50g
(2) 100 g
(3) 60 g
(4) 40 g
Show Answer
Answer: (4)
Solution
Let m gram of ice is added. From principal of calorimeter, heat gained (by ice) = heat lost (by water)
$20 \times 2.1 \times m + (m - 20) \times 334 = 50 \times 4.2 \times 40$
$376m = 8400 + 6680$
$m = 40.1$
$m \approx 40$ g
BETA
