Question

7. What mass of steam initially at 130°C is needed to warm 200 g of water in a 100-g
glass container from 20.0°C to 50.0°C? Specific heat of container is 837J/K9°C
Specific heat of steam is 2110J/Kg C
Specific heat of water is 4186J/K9°C
Latent heat of vapourisation is 2.26 10 J/Kg
a. 16.9 g
b. 14.9 g
c. 10.9 g
d. 12.9 g​

Answer 1

Heat lost by steam = heat gained by water + heat gained by container

m(s).L +m(s).c(s).(130–100) = m(w). c(w). 30 + m(c).c(c).30

You can look up L for steam, c(s), c(w) and when you know the container material, you can look up c(c) as well.