Answer
(a) The temperature of the mercury increases by $35.7~C^{\circ}$
(b) $2990~J$ of heat energy must be added to the water.
Work Step by Step
(a) The heat energy $Q$ required to change the temperature of a substance is:
$Q = m~c~\Delta T$
We can find the increase in temperature of the mercury as:
$Q = m~c~\Delta T$
$\Delta T = \frac{Q}{m~c}$
$\Delta T = \frac{100~J}{(0.020~kg)(140~J/kg~C^{\circ})}$
$\Delta T = 35.7~C^{\circ}$
The temperature of the mercury increases by $35.7~C^{\circ}$.
(b) We can find the heat energy required to raise the temperature of 20 g of water by the same amount.
$Q = m~c~\Delta T$
$Q = (0.020~kg)(4186~J/kg~C^{\circ})(35.7~C^{\circ})$
$Q = 2990~J$
$2990~J$ of heat energy must be added to the water.
