How much heat energy must be removed to cool a room from 70 °F to 40 °F, given a specific heat of 0.24 BTU/lb/°F and a specific volume of 12.7 cu ft/lb?

To determine the amount of heat energy that must be removed to cool a room from 70 °F to 40 °F, we can use the formula for calculating heat transfer, which is:

[ Q = m \times c \times \Delta T ]

where:

• ( Q ) is the heat energy (in BTU),

• ( m ) is the mass (in pounds),

• ( c ) is the specific heat (in BTU/lb/°F), and

• ( \Delta T ) is the change in temperature (in °F).

First, we need to calculate the change in temperature (( \Delta T )):

[ \Delta T = T_{initial} - T_{final} = 70 °F - 40 °F = 30 °F ]

Now, we need to find the mass of the air in the room. This requires knowing the volume of the room. Although the volume of the room is not provided directly in this question, we assume a hypothetical room size to illustrate the calculation.

The mass can be calculated using the specific volume:

[ m = \frac{V}{specific\ volume} ]

For instance, if we consider a room