Question

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

Answer 1

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

Answer 2

answer is 32 degree centigrade