Question

How much heat (in joules) is required to raise the temperature of 34.0 kg of water from \(15{\rm{\circ C}}\)to \(95{\rm{\circ C}}\)

Short answer

The heat required to raise the temperature of the given amount of water is \(1.14 \times {107}\,{\rm{J}}\).

Step-by-step solution

Definition of heat capacity

The heat capacity of a system represents the heat energy required to raise the temperature of the system. It depends on the mass and temperature of the system.

In other words, the mass and temperature of the system are directly proportional to the heat capacity of the system.

The expression for the heat generated by the water heater per hour is

\(Q = mc\left( {{T_{\rm{f}}} - {T_{\rm{i}}}} \right)\).

Here, \(m\) is mass of water, \(c\) is the specific heat capacity of water, \({T_{\rm{f}}}\) is the final temperature of water, and \({T_{\rm{i}}}\) is the initial temperature of water.

Given information

The mass of the water is \(m = 34.0\,{\rm{kg}}\).

The initial temperature of the water is \({T_{\rm{i}}} = 15\,{\rm{\circ C}}\).

The final temperature of the water is \({T_{\rm{f}}} = 95\,{\rm{\circ C}}\).

Calculation of the heat required to raise the temperature of the given amount of water

The heat required to raise the temperature of the given amount of water is calculated below:

\(\begin{array}{c}Q = \left( {34.0\,{\rm{kg}}} \right)\left( {4182\,\,{{\rm{J}} \mathord{\left/{\vphantom {{\rm{J}} {{\rm{kg}} \cdot {\rm{\circ C}}}}} \right.\\} {{\rm{kg}} \cdot {\rm{\circ C}}}}} \right)\left( {95{\rm{\circ C}} - 15\,{\rm{\circ C}}} \right)\\ = 11375040\,{\rm{J}}\\{\rm{ = 1}}{\rm{.14}} \times {\rm{1}}{{\rm{0}}7}\,{\rm{J}}\end{array}\)

Hence, the required heat\(1.14 \times {107}\,{\rm{J}}\).