5 g of ice at 0 ∘ C is dropped in a beaker containing 20 g of water at 40 ∘ C . The final temperature will be
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Heat neede by 5g of ice to get converted to water at 0°C is guven by
Q=mL
Q=5×80=400 calories
And heat that 20g ice can liberate
is given by
^
Q=mc/_\T
where delta T is change in temperature
Q=20×1×40
Q=800 calories
I.e, Even after melting all the ice 400 calories are still left that cause increse in final temperature of water
New mass of water is given by =5+20=25g
Q=400=25×1×(delta T )
8=delta T
Delta T =8
T°C-0 =8
T=8°C
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answer is 32 degree centigrade
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