A 0.510 g sample of steam at 105.4 °C is condensed into a container with 4.62 g of water at 14.2 °C . What is the final temperature of the water mixture if no heat is lost?

Answer 1

q = mC∆T

q = heat

m = mass

C = specific heat

∆T = change in temp

heat lost by steam must equal heat gained by water (conservation of energy)

heat lost by steam = (0.510 g)(1.89 J/gº)(105.4 - Tf) + (0.510 g)(2260 J/g) = 101.6 - 0.9639Tf

heat gained by water = (4.62 g)(4.184 J/gº)(tf - 14.2) = 19.3Tf - 274.5

101.6 - 0.9639Tf = 19.3Tf - 274.5

20.26Tf = 376.1

Tf = 18.6ºC