The concrete roof of a house of thickness 20 cm has an area 200 m². The temperature inside the house is 15 °C and outside is 35°C. Find the rate at which thermal energy will be conducted through the roof. The value of k for concrete is 0.65 Wm-'K (13000 JS")
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Answers
Answer:
⌬⠀Conduction of heat Energy , k = 0.65 W/m.k ,
⠀⠀⠀⌬⠀ Temperature Inside the House , T₁ = 15⁰ C or 288 K
⠀⠀⠀⌬⠀Temperature Outside the house , T₂ , a = 35⁰ C or 308 K
⠀⠀⠀⌬⠀Area , A = 200 m²
⠀⠀⠀⌬⠀ Thickness of roof , d = 20 cm or 0.2 m
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀★ According To The Question :
✇ The Rate of Conduction of Heat is Given by —
\begin{gathered}\\\qquad \star \:\:\pmb{\underline {\boxed {\sf{ \:\dfrac{Q}{T} \:=\: \Bigg\lgroup \: \dfrac{ k \:.\:A \:. \big( T_2 - T_1 \big)}{ d}\:\Bigg\rgroup}}}}\:\\\\\\\end{gathered}
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T
Q
=
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d
k.A.(T
2
−T
1
)
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T
Q
=
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d
k.A.(T
2
−T
1
)
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Where ,
⠀⠀⠀⠀
Q/T is the Rate of Conduction of Heat ,
k is the Conduction of Heat Energy ,
A is the Area ,
d is the Thickness ,
T₁ is the Temperature inside &
T₂ is the Temperature outside .
\begin{gathered}\\\qquad \dag\:\frak{\underline { Substituting \:known \:Values \:in \:Given \:Formula \:\::\:}}\\\\\end{gathered}
†
SubstitutingknownValuesinGivenFormula:
\begin{gathered} \dashrightarrow \sf \:\dfrac{Q}{T} \:= \: \dfrac{ k \:.\:A \:. \big( T_2 - T_1 \big)}{ d} \\\\\\\\ \dashrightarrow \sf \:\dfrac{Q}{T} \:= \: \dfrac{ 0.65 \:\times\:200 \: \big( 308 - 288 \big)}{ 0.2} \\\\\\\\ \dashrightarrow \sf \:\dfrac{Q}{T} \:= \: \dfrac{ 0.65 \:\times\:200 \: \big( 308 - 288 \big)}{ 0.2} \\\\\\\\ \dashrightarrow \sf \:\dfrac{Q}{T} \:= \: \dfrac{ 130 \: \big( 308 - 288 \big)}{ 0.2} \\\\\\\\ \dashrightarrow \sf \:\dfrac{Q}{T} \:= \: \dfrac{ 130 \: \times \: 20\:}{ 0.2} \\\\\\\\ \dashrightarrow \sf \:\dfrac{Q}{T} \:= \: \dfrac{ \:2600\:}{ 0.2} \\\\\\\\\dashrightarrow \sf \:\dfrac{Q}{T} \:= \: 13000\: W \: \\\\\\\\ \dashrightarrow \sf \:\dfrac{Q}{T} \:= \: 13000\: J.s \: \qquad \because \:\bigg\lgroup \: 1\:W \: = \: 1 \: J.s\:\bigg\rgroup\\\\\\\\ \dashrightarrow \pmb {\underline {\boxed {\purple {\:\frak{ \:\dfrac{Q}{T} \:= \:13,000\:J.s\:}}}}}\:\bigstar \: \\\\\\\end{gathered}
⇢
T
Q
=
d
k.A.(T
2
−T
1
)
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T
Q
=
0.2
0.65×200(308−288)
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T
Q
=
0.2
0.65×200(308−288)
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T
Q
=
0.2
130(308−288)
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T
Q
=
0.2
130×20
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T
Q
=
0.2
2600
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T
Q
=13000W
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T
Q
=13000J.s∵
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1W=1J.s
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T
Q
=13,000J.s
T
Q
=13,000J.s
★
∴
Explanation:
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