## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

(a) The temperature of the mercury increases by $35.7~C^{\circ}$ (b) $2990~J$ of heat energy must be added to the water.
(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.