Calculating the Cheaper Option for Heating Room

[pcoppin]

Homework Statement

An oil-filled radiator is used to maintain the temperature of a room at 21C when the outside temperature is 10C. The radiator has a surface area of 1.5 m2, and a surface temperature of 70C. Given that the radiator is made of a metal with a thermal conductivity 80 Wm-1K-1, thermal convection coefficient of 9.5 Wm-2K-1 and an emissivity of 0.8.

Question 1: If it was decided that a heat pump would be a better wat to heat this room, how much work would be done by the pump with a coefficient of performance of 9?

Question 2: If both the radiator and pump were run by electricity, which would be the cheaper option? Why?

Homework Equations

Currently, the rate at which heat is transferred into the room by convection plus radiation from the radiator is equal to the rate at which heat is conducted through the metal.
The Rate of Transfer is 1131.67 W (not sure if this is relevant)

The Attempt at a Solution

Not sure what to do as only given the coefficient of performance as 9.

 

[Andrew Mason]

Homework Statement

An oil-filled radiator is used to maintain the temperature of a room at 21C when the outside temperature is 10C. The radiator has a surface area of 1.5 m2, and a surface temperature of 70C. Given that the radiator is made of a metal with a thermal conductivity 80 Wm-1K-1, thermal convection coefficient of 9.5 Wm-2K-1 and an emissivity of 0.8.

Question 1: If it was decided that a heat pump would be a better wat to heat this room, how much work would be done by the pump with a coefficient of performance of 9?

Question 2: If both the radiator and pump were run by electricity, which would be the cheaper option? Why?

First, find the rate at which energy is radiated and convected away from the radiator surface. That is the rate of heat output. Then find the rate at which work is done by a heat pump with a COP of 9 that delivers that rate of heat output . Use:

$$COP = \dot Q_h/\dot W$$

AM

 

[Hamyrenz]

I think there is a formula in calculating the total amount of heat are you using. I don't remember. Where can I saw this formula? My suggestion is trying to apply the formula if you find it.