Final temp if no heat is lost

Question

A 0.515 g0.515 g sample of steam at 105.7 ∘C105.7⁢ ∘C is condensed into a container with 4.59 g4.59 g of water at 16.7 ∘C.16.7⁢ ∘C. What is the final temperature of the water mixture if no heat is lost? The specific heat of water is 4.18  J g⋅ ∘C,4.18  J g⋅ ∘C, the specific heat of steam is 2.01  J g⋅ ∘C,2.01  J g⋅ ∘C, and Δ𝐻vap=40.7 kJ/mol.

Answer 1

This problem is best approached in a step-wise fashion. We have 0.515 g of steam @ 105.7ºC being condensed into 4.59 g of water @ 16.7ºC. The total heat lost by the steam must be equal to the total heat gained by the water. The equation for heat is q = mC∆T when no phase change, and q = m∆H when there is a phase change. C = specific heat; m = mass and ∆T = change in temp

We will find heat lost by steam and set it equal to heat gained by water and solve for final temp (Tf)

Step1: heat lost by steam going from 105.7 to 100º: q = mC∆T = (0.515 g)(2.01 J/gº)(5.7º) = 5.90 J

Step2: heat lost by steam turning to water @ 100º: q = m∆Hvap = (0.515 g)(1 mol/18g)(40.7 kJ/mol) = 1.16 kJ

Step 3: heat lost by water going from 100 to Tf: q = mC∆T = (0.515 g)(4.18 J/gº)(100-Tf) = 215J/º - 2.15Tf

Step 4: total heat lost by steam = 5.90 + 1160 + 215 - 2.15Tf (NOTE: change 1.16 kJ to 1160 J)

Step 5: heat gained by water @ 16.7º: q = mC∆T = (4.59 g)(4.18 J/gº)(Tf - 16.7º) = 19.2Tf - 320

Step 6: set heat lost by steam = heat gained by water: 5.90 + 1160 + 215 - 2.15Tf = 19.2Tf - 320

Step 7: solve for Tf (final temperature)

5.90 + 1160 + 215 - 2.15Tf = 19.2Tf - 320

1381 - 2.15Tf = 19.2Tf - 320

21.4Tf = 1701

Tf = 79.5ºC

(be sure to check the math, and rounding)