A 2kg block of ice initially at T = −20 ◦C is put in contact with a very large amount of liquid water at T = 0 ◦C (by “very large amount” we imply that by the time ice and water reach thermal equilibrium, there will still be some liquid water left). Ice and water can exchange heat just with each other but not with the environment. The speciﬁc heat of water is 4186 J and of ice is 2050J.
2)amount of heat absorbed by the ice in the process
3)final mass of the ice at equilibrium
deltaEbond + deltaEthermal = Q + W
deltaEthermal = m•Csp•deltaT
The Attempt at a Solution
My first guess is that the equilibrium temperature is 0ºC, because we're to assume that water is still present.
assuming thats correct could I then use deltaEthermal = m•Csp•deltaT = 2kg • 2050J/kg•K • (0ºC-(-20ºC)) = 82kJ
and then, assuming that all the ice melts, calculate deltaEbond = |(mf - mi)•deltaH(melting, ice)| = |(0kg - 2kg)•6.01| = 12.02kJ
then, since Q=deltaEthermal + delta Ebond, Q = 82kJ + 12.02kJ = 96.02kJ
But I had to assume there was enough E to melt all the ice to calculate this... and one of the questions asks to calculate the amount of ice left!
I feel fairly stupid for not being able to do this, I think I'm waaaaaaaay over-complicating it. Thanks internet!!!!