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QJEEM19Ph1a48: JEE Main 2019 Physics - Class 11 JEEM19Ph1a48 Solution: Ice at –20 o C is added to 50 g of water at 40 o C. When the…

Ice at –20oC is added to 50 g of water at 40oC. When the temperature of the mixture reaches 0oC, 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 = 4.2J/g/oC Specific heat of Ice = 2.1J/g/oC Heat of fusion of water at 0oC= 334J/g)

Multiple Choice Options

  • A 100 g
  • B 60 g
  • C 50 g
  • D 40 g

Step-by-Step Verified Solution

Let amount of ice is m gm.

According to principal of calorimeter heat taken by ice = heat given by water

∴  20 × 2.1 × m + (m − 20) × 334

= 50 × 4.2 × 40

376 m = 8400 + 6680

m = 40.1

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