Homes may be heated by pumping hot water through radiators. What mass of water will provide the same amount of heat when cooled from 95.0 to 35.0 °C, as the heat provided when 100 g of steam is cooled from 110 °C to 100 °C.

mass of water is 7.43 g

Explanation:

q = mc∆T

qsteam = 100 g * 1.864 J / (g °C) * (110 - 100) °C = 1864 J

qwater = mwater * 4.184 J / (g °C) * (95 - 35) °C = mwater * 251.04 J/g = qsteam = 1864 J

mwater = 1864 J / 251.04 J/g = 7.43 g