a heat pump unit allows a 5% room temp drop before it starts back up. if the room temperature is 72 degrees, what does it drop to before the heat pump kicks in?

Question

I know the question uses direct variation to solve it, I'm just not sure how to set up the first calculation (the one before you cross multiply) 

Answer 1

We only allow the room temperature to drop 5% before the heat pump kicks in.

That means the heat pump kicks in when the temperature is at 95% of 72 degrees.

 

.95(72) = 68.4 degrees

Answer 2

I believe Andrew's answer is what the problem is looking for.

 

WARNING: the following discussion is about the difficulty in making math problem out of the real world.

 

But, pose the following question: If the room temperature were 0, then what would a 5% drop in temperature be? [.95(0) = 0 degrees]???

 

Either the 5% is accurate only for a limited range of temperatures (say 50-120 degrees Fahrenheit), or it should be based on the heat energy of the room in which case it should be based on an absolute zero-based scale like Kelvin.

 

For this problem we should either translate everything to Kelvin or use the Rankine scale (corollary to Kelvin for the Fahrenheit world)  where absolute zero is -459.67 (and the size of one degree is the same as in Fahrenheit).

 

Therefore the room temperature is 531.67 °R (72 °F) and .95(531.67) = 505.08 °R (or 45.42 °F) is where the heat pump would turn on.

 

If you read this I hope you read it with a smile.