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

Answer:

$75^{\circ}C$

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

$\Delta T=\frac{Q}{mC_s}=\frac{147000}{(700)(4.2)}=50^{\circ}C$

The initial temperature was

$T_1 = 25^{\circ}C$

So the final temperature will be

$T_2 = T_1 + \Delta T=25+50=75^{\circ}C$