A house thermostat is normally set to 22 degrees C, but at night it is turned down to 16 degrees...

Question:

A house thermostat is normally set to 22 degrees C but at night, it is turned down to 16 degrees C for 9.0 h. Estimate how much more heat would be needed (state as a percentage of daily usage) if the thermostat were NOT turned down at night. Assume that the outside temperature averages 0 degrees C for the 9.0 h at night and 8 degrees C for the remainder of the day, and that the heat loss from the house is proportional to the temperature difference inside and out. To obtain an estimate from the data, you must make other simplifying assumptions; state what these are.

Heat Transfer:

Heat transfer is the process by which thermal energy is transferred due to the variation in temperature of two bodies or medium. In nature, the transfer of thermal energy takes place from high temperatures to low temperatures, or vice-versa by doing some external work.

Answer and Explanation:


We are given the following data:

  • When thermostat is tuned down at night, {eq}\Delta {T_2} = 16^\circ {\rm{C}}{/eq}.
  • When thermostat is not turned down, {eq}\Delta {T_2} = 22^\circ {\rm{C}}{/eq}.
  • Temperature difference between house and outside air, {eq}\Delta {T_1} = = 22 - 8 = 14^\circ {\rm{C}}{/eq}.
  • Number of hours before turning down is, {eq}{t_1} = 24 - 9 = 15\;{\rm{h}}{/eq}.
  • Number of hours at night is, {eq}{t_2} = 9\;{\rm{h}}{/eq}.


The expression for rate of heat is:

{eq}\begin{align*} \dfrac{Q}{t} &= \Delta T\dfrac{{kA}}{l}\\ Q &= \alpha t\Delta T \end{align*}{/eq}, where:

  • {eq}Q{/eq} is heat.
  • {eq}\Delta T{/eq} is the change in temperature.
  • {eq}\alpha{/eq} is average heat from conducting surface.
  • A is area.
  • t is time.


The expression for required heat when thermostat is turning down is,

{eq}{Q_{td}} = \alpha \left[ {{t_1}\Delta {T_1} + {t_2}\Delta {T_2}} \right] {/eq}


Substituting the values into the above expression gives us:

{eq}\begin{align*} {Q_{td}} &= \alpha \left[ {15 \times 14 + 9 \times 16} \right]\\ {Q_{td}} &= 354\alpha \;{\rm{J}} \end{align*}{/eq}


The expression for required heat when thermostat is not turned down at night is:

{eq}{Q_{ntd}} = \alpha \left[ {{t_1}\Delta {T_1} + {t_2}\Delta {T_2}} \right]{/eq}


Substituting the values into above expression, we get:

{eq}\begin{align*} {Q_{td}} &= \alpha \left[ {15 \times 14 + 9 \times 22} \right]\\ {Q_{ntd}} &= 408\alpha \;{\rm{J}} \end{align*}{/eq}


The percentage more heat needed is:

{eq}\Delta Q = \dfrac{{{Q_{ntd}} - {Q_{td}}}}{{{Q_{td}}}} \times 100{/eq}


Substituting the values into the above expression, we have:

{eq}\begin{align*} \Delta Q &= \dfrac{{408\alpha - 354\alpha }}{{354\alpha }} \times 100\\ \Delta Q &= 15\% \end{align*}{/eq}


Thus, 15% more heat would be needed if the thermostat were NOT turned down at night.


Learn more about this topic:

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Thermal Expansion & Heat Transfer

from High School Physics: Help and Review

Chapter 17 / Lesson 12
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