The temperature of a sample of water increases from 30.0° C to 40.0°C as 6940 joules of heat are added. What is the mass of the sample of water?

Question

The partial answer is: 1XX g

Answer 1

q = mC∆T

q = heat = 6940 J

m = mass = ?

C = specific heat of water = 4.184 J/g/deg

∆T = change in temperature = 40.0 - 30.0 = 10.0ºC

6940 J = (m)(4.184J/g/deg)(10 deg)

m = 165.9 g = 166 g