Question

5. The specific heat capacity of water is 4.2 J/gºC. If 700.0 g of water absorbed 147000 J of thermal energy and its

temperature rose from 25°C, what temperature will it reach? Show your work

Answer 1

`75^(\circ)C`

Step-by-step explanation:

When an amount of energy Q is supplied to a substance of mass m, the temperature of the substance increases by
`\Delta T`, according to the equation:

`Q=mC_s \Delta T`

where

`C_s` is the specific heat capacity of the substance

Here we have

m = 700.0 g of water

`C_s = 4.2 J/gC` is the specific heat capacity of water

`Q=147000 J` is the energy supplied

Solving for
`\Delta T`, we find that the temperature of the water increases by

`\Delta T=(Q)/(mC_s)=(147000)/((700)(4.2))=50^(\circ)C`

The initial temperature was

`T_1 = 25^(\circ)C`

So the final temperature will be

`T_2 = T_1 + \Delta T=25+50=75^(\circ)C`