Question

A copper container of mass 0.080 kg and specific heat 387 Jkg-1K-1 contains 0.30 kg of water and 0.040 kg of ice at 0˚C. Steam at 100˚C is passed into the water and its temperature stabilizes at 20.0˚C. Find the mass of the water left in the container. Assume the system is insulated from its environment. --- Please have the answer be completely clear. There is another answer to this question but it is so hard to decipher. Please show Q term formulas with terms only and then with values.

Answer 1

mass of copper, m = 0.080 kg
specific heat of copper, C = 387
initial mass of weater = m' = 0.3 kg
initial mass of ice M' = 0.04 kg
initial temp, Ti = 0 C
Final temperature = Tf = 20 C
mass of steam condensed = M
initial temp pf steam, T = 100 C
latent heat of vapourisation of water, L = 2264760 J / kg C
latent heat of fusion of water, l = 334000 J / kg C
heat capacity of water, c = 4184 J / kg C

So heat lost by steam = ML + Mc(T - Tf) = M(2264760 + 4184*(100-20)) = 2599480M J
heat gained by water = m'*c(Tf - Ti) = 0.3*4184*20 = 25104 J
heat gained by copper = mC(Tf - Ti) = 0.08*387(20) = 619.2 J
heat gained by ice = M'l + M'c(Tf - Ti) = 0.04(334000 + 4184*20) = 16707.2 J
heat gained = heat lost
25104+619.2+16707.2 = 42430.4 = 2599480M
M = 16.322 grams
mass of water left = M + M' + m = 0.3563 kg