Ice at –20° C is added to 50g of water at $40° C$ , when the temperature of the mixture reaches $0° C$ , it is formed 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 $= 4.2$ /J/g/°C
Heat of fusion of water at 0°C = 334J/g)

Question

Ice at –20° C is added to 50g of water at

40° C

, when the temperature of the mixture reaches

0° C

, it is formed 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

= 4.2

/J/g/°C
Heat of fusion of water at 0°C = 334J/g)

Accepted Answer

Correct answer: D. (d) JEE Main 2019 Let m gram of ice is added. From principle of calorimeter Heat gained (by ice) = heat lost (by water) ∴20 2.1 m+(m-20) 334 =50 4.2 40 376m=8400+6680 m=40.1

Solution

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