Question

How many joules of heat are absorbed to raise the temperature of 435 grams of water at 1 atm from 25°c to its boiling point?

Answer 1

Answer: 136.6 kJ  

Explanation:

The amount of heat $Q$ absorbed in the temperature variation of a material is:

$Q=m. c. \Delta T$   (1)

Where:

$m=435g$  is the mass  of water

$c$  is the specific heat of the element. In the case of water $c=1 cal/g\°C$

$\Delta T$  is the variation in temperature, which in this case is  $\Delta T=100\°C-25\°C=75\°C$  

(The boiling point of water at the pressure of 1 atm is  $100\°C$)

Rewriting equation (1) with the known values:

$Q=(435g)(1 cal/g\°C)(75\°C)$  

(2)

$Q=32625 cal$   (3)

Nevertheless, we are asked to find this value in Joules. So, we have to convert this 32625 calories to Joules, knowing the following:

$1 cal=4.187 J$   (4)

Hence:

$Q=32625 cal=136600.875 J=136.6kJ$