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Q56PE
Expert-verified(a) Cherry-red embers in a fireplace are at 850ºC and have an exposed area of \({\bf{0}}{\bf{.200}}\;{{\bf{m}}^{\bf{2}}}\) and an emissivity of 0.980. The surrounding room has a temperature of 18.0ºC. If 50% of the radiant energy enters the room, what is the net rate of radiant heat transfer in kilowatts? (b) Does your answer support the contention that most of the heat transfer into a room by a fireplace comes from infrared radiation?
The net rate of radiant heat transfer is \({\bf{8}}{\bf{.8}}\;{\bf{kW}}\).
(b) The most of heat transfer into room by a fireplace comes from infrared radiation.
The area of fireplace is \(A = 0.200\;{{\rm{m}}^2}\)
The temperature of fireplace is \({T_1} = 850^\circ {\rm{C}} = 1123\;{\rm{K}}\)
The temperature of surrounding room is \({T_2} = 18^\circ {\rm{C}} = 291\;{\rm{K}}\)
The emissivity of fireplace is \(\varepsilon = 0.980\)
The percentage of radiant energy enters to room is \(x = 50\% = 0.5\)
The radiant energy transfer is calculated by using the Stefan law. This law gives the energy transfer without any medium.
The net rate of radiant heat transfer is given as:
\(Q = x\varepsilon \sigma A\left( {T_2^4 - T_1^4} \right)\)
Here, \(\sigma \) is Stefan’s constant and its value is \(5.67 \times {10^{ - 8}}\;{\rm{W}} \cdot {{\rm{m}}^{ - 2}} \cdot {{\rm{K}}^4}\)
\(\begin{aligned}{}Q &= \left( {0.5} \right)\left( {0.980} \right)\left( {5.67 \times {{10}^{ - 8}}\;{\rm{W}} \cdot {{\rm{m}}^{ - 2}} \cdot {{\rm{K}}^4}} \right)\left( {0.200\;{{\rm{m}}^2}} \right)\left( {{{\left( {1123\;{\rm{K}}} \right)}^4} - {{\left( {291\;{\rm{K}}} \right)}^4}} \right)\\Q &= 8.798 \times {10^3}\;{\rm{W}}\\Q& = \left( {8.798 \times {{10}^3}\;{\rm{W}}} \right)\left( {\frac{{1\;{\rm{kW}}}}{{{{10}^3}\;{\rm{W}}}}} \right)\\Q \approx 8.8\;{\rm{kW}}\end{aligned}\)
Therefore, the net rate of radiant heat transfer is \(8.8\;{\rm{kW}}\).
(b)
The heat transfer into room by a fireplace comes from infrared radiation because infrared radiation does not requires any medium for heat transfer.
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