How much heat is required to warm 1.70 l of water from 30.0 ∘c to 100.0 ∘c? (assume a density of 1.0g/ml for the water.) ?

Quantity of heat = mass x specific heat capacity x change in temp.

Q = m * c * change in Temperature

Mass = Density * Volume = 1g/mL * 1.7L * 1000mL/1 = 1700g

Q=1700 * 4.18 * (100.0â’30.0)

Q = 497420 J

Heat absorbed is calculated by multiplying the heat capacity of water by mass and the by change in temperature. The heat capacity of water is 4184 J/kg/C.

Thus, heat = mcθ

The mass of water will be given by (1700 * 1)

Thus the heat absorbed will be (1700/1000) * 4184 * 70

= 497896 J

= 497.896 kJ