Question

A copper calorimeter can with mass 0.860 kg contains 0.185 kg of water and 0.020 kg of ice in thermal equilibrium at atmospheric pressure.

Part A:

What is the temperature of the ice–water mixture?

Part B:

If 0.775 kg of lead at a temperature of 255 ∘C is dropped into the can, what is the final temperature of the system? (Assume no heat is lost to the surroundings.)

Answer 1

(a) At atmospheric level, the water and ice can only remain at equilibrium at 273K.So, the temperature of the ice -water mixture is 273K.

(b) 0,775 kg lead at 528K dropped into the can. There will be heat exchange and heat lost by lead will be heat gained by copper can and ice -water mixture.

So, first check it has enough energy to overcome the latent heat of fusion, as first ice will convert to water.

latent heat of fusion for 0.02kg ice is 6.68KJ

specific heat of lead is 0.16

specific heat of ice is 2.03

specific heat of copper is 0.385

specific heat of water is 4.18

max heat lead can lost is 0.775*0.16*(528-273)=31.62KJ

So, all ice will fuse and by heat balance, assuming the system will reach a final temperature T.

0.775*0.16*(528 - T)=6.68+0.185*(T-273)*4.18 +0.02*(T-273)*4.18 + 0.86*0.385*(T-273)

T=322.82K