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

Step-by-step 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`