Question

How much heat is required to warm 1.70 L of water from 26.0 ∘C to 100.0 ∘C? (Assume a density of 1.0g/mL for the water.)

Answer 1

To find out the heat required we will use the equation below:

q = Cp x m x T

means; Heat = specific heat x mass x change in temp.

Given :   T = 100 - 26 = 74^oC

                 m = 1.70 L = 1700 mL = mass of 1700 g

                 Cp for water = 4.184 J/g-deg C

So q = 4.184 J/gdegree C x 1700g x 74 degree C

        = 526347.2 Joules or 5.26 x 10⁵ J

( The degree C and grams cancel out here to give you Joules as unit of answer. )