How much heat is required to warm 1.70 l of water from 22.0 ∘c to 100.0 ∘c? (assume a density of 1.0g/ml for the water?

Mass of thr water = 1.7 x (1000 mL / 1 L) * (1 g / 1 mL) = 1700 grams

T1 = 22 Celsius; T2 = 100 Celsius

dT = 100 - 22 = 78 Celsius

Specific heat of water C = 4.186 J/gm

We have heat energy Q (heat) = C x m x dT = 4.186 x 1700 x 78 = 555063.6 J

So the heat required Q = 555KJ Approx

Q (heat) = MC delta T

M=mass

c = specific heat capacity

delta T = change in temperature

M=density x volume which is = 1700ml x 1.0g/ml = 1700g

change in temperature = 373k-295k=78 K

specific heat capacity for water is 4.186 j/g

heat is therefore = 1700g x 4.186j/g x 78k=555063.6j or 555.0636Kj