Question

A heat pump in cooling mode has a SEER* (Seasonal Energy Efficiency Ratio) rating of 16 BTU/(W⋅h). Calculate the amount of work required to remove 2,282 J of energy from inside a business.

Answer 1

The amount of work required to remove 2,282 J of energy from inside a business using a heat pump with a SEER rating of 16 BTU/(W⋅h) is approximately 0.135 W⋅h.

Step-by-step explanation:

The amount of work required to remove 2,282 J of energy from inside a business using a heat pump with a SEER rating of 16 BTU/(W⋅h) can be calculated by converting the energy into BTUs and dividing it by the SEER rating.

First, we convert 2,282 J to BTUs:

• 1 J = 0.000947817 BTU
• 2,282 J = 2,282 * 0.000947817 = 2.167 BTU

Then, we divide the BTU value by the SEER rating to find the amount of work required:

• Work (in W⋅h) = 2.167 BTU / 16 BTU/(W⋅h) = 0.135 W⋅h

Therefore, the amount of work required to remove 2,282 J of energy from inside a business is approximately 0.135 W⋅h.