JEE MAIN - Physics (2019 - 11th January Morning Slot - No. 26)
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)
100 g
60 g
50 g
40 g
Explanation
Let amount of ice is m gm.
According to principal of calorimeter heat taken by ice = heat given by water
$$ \therefore $$ 20 $$ \times $$ 2.1 $$ \times $$ m + (m $$-$$ 20) $$ \times $$ 334
= 50 $$ \times $$ 4.2 $$ \times $$ 40
376 m = 8400 + 6680
m = 40.1
According to principal of calorimeter heat taken by ice = heat given by water
$$ \therefore $$ 20 $$ \times $$ 2.1 $$ \times $$ m + (m $$-$$ 20) $$ \times $$ 334
= 50 $$ \times $$ 4.2 $$ \times $$ 40
376 m = 8400 + 6680
m = 40.1
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