How much heat is needed to increase the temperature of 20 lb of water from 32°F to 212°F?

How much heat is needed to increase the temperature of 20 lb of water from 32°F to 212°F?

• A. 1,800 BTU
• B. 2,800 BTU
• C. 3,600 BTU
• D. 4,600 BTU

Answer: (C)

To determine the amount of heat needed to increase the temperature of a certain mass of water, you can use the formula: \[ Q = m \times C \times \Delta T \] where: - \( Q \) is the heat energy in BTUs, - \( m \) is the mass of the water in pounds, - \( C \) is the specific heat capacity of water (approximately 1 BTU/lb°F), - \( \Delta T \) is the change in temperature in degrees Fahrenheit. In this case, the mass of water is 20 lb, the initial temperature is 32°F, and the final temperature is 212°F. First, we need to calculate the change in temperature (\( \Delta T \)): \[ \Delta T = 212°F - 32°F = 180°F \] Now, substitute the values into the formula: \[ Q = 20 \, \text{lb} \times 1 \, \text{BTU/lb°F} \times 180°F \] Calculating that gives: \[ Q = 20 \times 180 = 3600 \, \text{BTU} \] Thus, the amount of heat required to

To determine the amount of heat needed to increase the temperature of a certain mass of water, you can use the formula:

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

where:

• ( Q ) is the heat energy in BTUs,

• ( m ) is the mass of the water in pounds,

• ( C ) is the specific heat capacity of water (approximately 1 BTU/lb°F),

• ( \Delta T ) is the change in temperature in degrees Fahrenheit.

In this case, the mass of water is 20 lb, the initial temperature is 32°F, and the final temperature is 212°F. First, we need to calculate the change in temperature (( \Delta T )):

[ \Delta T = 212°F - 32°F = 180°F ]

Now, substitute the values into the formula:

[ Q = 20 , \text{lb} \times 1 , \text{BTU/lb°F} \times 180°F ]

Calculating that gives:

[ Q = 20 \times 180 = 3600 , \text{BTU} ]

Thus, the amount of heat required to