Question

Homes may be heated by pumping hot water through radiators. What mass of water (in g) will provide the same amount of heat when cooled from 90.7 to 39.4°C, as the heat provided when 182 g of steam is cooled from 118°C to 100.°C. (Assume that the specific heat of liquid water is 4.187 J/g·°C over the given temperature range. Also assume that the specific heat of water vapor above its boiling point is 2.078 J/g·°C.)

Answer 1

Answer:

a mass of water required is mw= 1273.26 gr = 1.27376 Kg

Explanation:

Assuming that the steam also gives out latent heat, the heat provided should be same for cooling the hot water than cooling the steam and condense it completely:

Q = mw * cw * ΔTw = ms * cs * ΔTw + ms * L

where m = mass , c= specific heat , ΔT=temperature change, L = latent heat of condensation

therefore

mw = ( ms * cs * ΔTw + ms * L )/ (cw * ΔTw )

replacing values

mw = [182g * 2.078 J/g°C*(118°C-100°C) + 118 g * 2260 J/g ] /[4.187 J/g°C * (90.7°C-39.4°C)] = 1273.26 gr = 1.27376 Kg