Question

Suppose that 20.0g of ice at 32.0 Farenheit, and 0.150 L of water at 60.0 Farenheit are mixed ideally so that energy is exchanged only between them. What is the equilibrium temperature?

Answer 1

32 farenheit=0 degree celcius

60 farenheit=15.5556 degree celcius

initial mass of water=volume*density=0.15 L*(1 kg/L)=0.15 kg

let final temperature be T degree celcius.

then heat lost by water=heat gained by ice

==>mass of water*specific heat of water*temperature change=mass of ice*latent heat of fusion of ice+mass of melted water*specific heat of water*temperature change

==>0.15*4186*(15.5556-T)=0.02*334*1000+0.02*4186*(T-0)

==>0.15*4186*15.5556-0.02*334*1000=T*(0.02*4186+0.15*4186)

==>T=4.3385 degree celcius

=39.8093 Farenheit