Question

9.1 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")​

Answer 1

Answer:

Rate of conduction of heat is given by the formula:

$\boxed{ \bf{\dfrac{Q}{t} = \dfrac{k.A. \Delta T}{l}}}$

where, Q/t refers rate of conduction of heat energy, k refers to conductivity of the material, A refers to the Area, ΔT refers to change in temperature and l refers to the thickness.

According to the question,

• k = 0.65 W/m.K
• Temperature outside = 35° C = 308 K
• Temperature inside = 15° C = 288 K
• Area = 200 m²
• Thickness = 20 cm = 0.2 m

Substituting the given information we get:

$\implies \dfrac{Q}{t} = \dfrac{ 0.65 \times 200 \times (308 - 288)}{0.2}\\\\\\\implies \dfrac{Q}{t} = \dfrac{130 \times 20}{0.2}\\\\\\\implies \boxed{\bf{\dfrac{Q}{t} = 13000\:\:W}}$

Since 1 W = 1 J.s, we get the rate of heat conducted to be 13,000 J.s.

Answer 2

⠀⠀⠀⌬⠀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 —

$\\\qquad \star \:\:\pmb{\underline {\boxed {\sf{ \:\dfrac{Q}{T} \:=\: \Bigg\lgroup \: \dfrac{ k \:.\:A \:. \big( T_2 - T_1 \big)}{ d}\:\Bigg\rgroup}}}}\:$

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 .

$\\\qquad \dag\:\frak{\underline { Substituting \:known \:Values \:in \:Given \:Formula \:\::\:}}$

$\dashrightarrow \sf \:\dfrac{Q}{T} \:= \: \dfrac{ k \:.\:A \:. \big( T_2 - T_1 \big)}{ d} \\\\\\\\ \dashrightarrow \sf \:\dfrac{Q}{T} \:= \: \dfrac{ 0.65 \:\times\:200 \: \big( 308 - 288 \big)}{ 0.2} \\\\\\\\ \dashrightarrow \sf \:\dfrac{Q}{T} \:= \: \dfrac{ 0.65 \:\times\:200 \: \big( 308 - 288 \big)}{ 0.2} \\\\\\\\ \dashrightarrow \sf \:\dfrac{Q}{T} \:= \: \dfrac{ 130 \: \big( 308 - 288 \big)}{ 0.2} \\\\\\\\ \dashrightarrow \sf \:\dfrac{Q}{T} \:= \: \dfrac{ 130 \: \times \: 20\:}{ 0.2} \\\\\\\\ \dashrightarrow \sf \:\dfrac{Q}{T} \:= \: \dfrac{ \:2600\:}{ 0.2} \\\\\\\\\dashrightarrow \sf \:\dfrac{Q}{T} \:= \: 13000\: W \: \\\\\\\\ \dashrightarrow \sf \:\dfrac{Q}{T} \:= \: 13000\: J.s \: \qquad \because \:\bigg\lgroup \: 1\:W \: = \: 1 \: J.s\:\bigg\rgroup\\\\\\\\ \dashrightarrow \pmb {\underline {\boxed {\purple {\:\frak{ \:\dfrac{Q}{T} \:= \:13,000\:J.s\:}}}}}\:\bigstar \:$

∴ Hence, The rate at which thermal energy will be conducted is 13,000 J.s .