Question

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?

Answer 1

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

Answer 2

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