Question

The temperature very near a glass window pane on the inside of a house is measured as 10∘C and the temperature on the outside near the window is measured as 4∘C. The window has dimensions 1.3 m by 0.8 m. The thickness d of the window is 5 mm, and its thermal conductivity k is 0.80 Js⋅m⋅∘C. Find the rate at which energy is transferred from the inside of the room to the outside environment. *

Answer 1

The rate at which energy is transferred from the inside of the room to the outside environment is 998.4 W/m²

The rate of energy transfer

The rate of energy transfer through the glass window pane is given by

P = kA(T₂ - T₁)/d where

• k = thermal conductivity = 0.80 Js.m.⁰C,
• A = area of window = 1.3 m × 0.8 m = 1.04 m²,
• T₁ = outside temperature = 4 ⁰C,
• T₂ = inside temperature = 10 ⁰C and
• d = thickness of window = 5mm = 0.005 m

So, substituting the values of the variables into the equation, we have

P = kA(T₂ - T₁)/d

P = 0.80 Js/m-⁰C × 1.04 m²(10 ⁰C - 4 ⁰C)/0.005 m

P = 0.832 Js/m³-⁰C × 6 ⁰C/0.005 m

P = 4.992 Js.m³/0.005 m

P = 998.4 Js/m²

P = 998.4 W/m²

So, the rate at which energy is transferred from the inside of the room to the outside environment is 998.4 W/m²

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