An aluminium pot has a mass of 200g and contains 400g of ice at 0°C. How much heat would be needed to melt that ice and then raise the temperature of the resulting water to 20°C. The specific heat capacity of water and aluminium are 4200 J Kg ^-1 and 910 J Kg ^-1 respectively, and the latent heat of fusion of ice is 330 KJ/Kg.

My attempt so far is as follows:

Heat from ice to water - Q = mL = .4Kg x 330x10^3J = 132x10^3J

Heat for water from (0°C to 20°C) - Q = mcΔT = .4Kg x 4200 x (20-0) = 33.6x10^3J

Heat for pot from (0°C to 20°C) - Q = mcΔT = .2Kg x 910 x (20-0) = 3.64x10^3J

Thus the answer is 3.64x10^3J + 33.6x10^3J + 132x10^3J = 169.24x10^3J

I don't know how to do this question but would try go this way about it, however it looks terribly wrong to me and would like some help please.


closed as off-topic by David Z Jan 11 '14 at 23:36

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  • $\begingroup$ Why does it look terribly wrong? Your $0-20$'s should be $20-0's (minus, not to) The message that the big number is melting the ice is correct. $\endgroup$ – Ross Millikan Jan 11 '14 at 23:39
  • $\begingroup$ Thanks for the feedback. I guess it looks wrong because I don't know the correct way to approach the question and I just doubted my attempt. Is the rest incorrect? @Ross Millikan $\endgroup$ – user37216 Jan 11 '14 at 23:51
  • $\begingroup$ You approached exactly correctly. I didn't check the numbers too carefully, but they look right $\endgroup$ – Ross Millikan Jan 12 '14 at 0:16
  • $\begingroup$ Oh, that's great news. As long as the method was correct. thanks a lot for the help bro. $\endgroup$ – user37216 Jan 12 '14 at 2:15