<p>In winter, a house loses heat at a rate of L(T-To) where T is the temperature

Question

<p>In winter, a house loses heat at a rate of L(T-To) where T is the temperature inside the house, To is the temperature outside the house, and L is a positive constant. (A) suppose we want to keep the temperature inside constant by hooking up and ideal heat pump (a refrigerator) that transfers heat from the outside to the inside. If the power of the ideal heat pump is W, show that the steady state temperature is</p>
<p>T=To + (1 + sqrt(1 + 4LTo/W))(W/2L).</p>
<p>(B) If we just use a space heater which simply converts all input power to heat, what is the steady state temperature?</p>

Explanation / Answer

let rate of heat support by the heater be Q Q=W*T/(T-T0). Q=L(T-T0). so that W*T/(T-T0)=L(T-T0). let T-T0=X. so W*(X+T0)=L*(T-T0)^2. so W*X+WT0=L*X^2. solve this we have . X=(1 + sqrt(1 + 4LTo/W))(W/2L) so we have T= T0+(1 + sqrt(1 + 4LTo/W))(W/2L) ------------- we will have that. L(T-T0)=W so T=W/L+T0

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