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 100g of steam is cooled rom 110c to 100c.

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

8.08 grams

Explanation:

Given;

Mass of the steam, m = 100 g

Change in temperature of water = (95°C - 35°C) = 60°C

Change in temperature of steam = (110°C - 100°C) = 10°C

let the mass of water by 'M'

also,

We know

specific heat of water, c = 4.184 J/Kg.°C

And, specific heat of steam, C = 2.03 J/Kg.°C

Now,

Heat released = Mass × specific heat × change in temperature

thus, for the condition given in the question

M × 4.184 × 60 = 100 × 2.03 × 10

or

M = 8.08 grams