REFER TO THE PROBLEM BELOW: How much work must the pump do to deliver 3350 J of heat into the house on another day when the outdoor temperature is 252 K (21 °C)? An ideal, or Carnot, heat pump is used to heat a house to a temperature of 294 K (21 °C). How much work must the pump do to deliver 3350 J of heat into the house on a day when the outdoor temperature is 273 K (0 °C)? (Cutnell, 2009)
REFER TO THE PROBLEM BELOW: How much work must the pump do to deliver 3350 J of heat into the house on another day when the outdoor temperature is 252 K (21 °C)? An ideal, or Carnot, heat pump is used to heat a house to a temperature of 294 K (21 °C). How much work must the pump do to deliver 3350 J of heat into the house on a day when the outdoor temperature is 273 K (0 °C)? (Cutnell, 2009)
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Transcribed Image Text:REFER TO THE PROBLEM BELOW: How much work must the pump do to deliver 3350 J of
heat into the house on another day when the outdoor temperature is 252 K (21 °C)?
An ideal, or Carnot, heat pump is used to heat a house to a temperature of 294 K (21 °C).
How much work must the pump do to deliver 3350 J of heat into the house on a day when
the outdoor temperature is 273 K (0 °C)? (Cutnell, 2009)
Given:
TH=294 K
Tc=273 K
|QH| = 3350 J
Required: Won 2
= ?
Solution:
Using equation 1, QH=Won +Qcl, we find the unknown
magnitude of work which is equal to
|Won = |QH|-|Ocl-
To find Qc, we will use equation 2 and the information given about
the temperatures TH (for the hot reservoir) and Te (for the cold
reservoir).
|oc|
|QH|
7
1ocl = 19 (7)
Substituting the magnitude of Qe to the equation for work, we have
|Won| = |QH|-|QH| ()
|Won = Q(1-())
Therefore, the magnitude of work for the given temperature is
(for 273 K):
|Won| = |QH|(1-())
273
|Won (3350 J)(1-(K)) = 239 J
294
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