calculate the rate of loss of heat through a glass window of area 1000cm^2 and thickness 0.4 cm when temperature inside is 37degreeC and outside is 5 degree C .coefficient of thermal conductivity of glass 2.2×10^-3 cals/cm/kg.
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Answered by
98
use formula ,
dQ/ dt = KAdT/dL
where
dQ/dt is rate of change of heat , K is thermal conductivity , A is cross-section area , dT is the temperature difference , and dL is thickness of wire .
now,
dQ/dt = 2.2 × 10^-3 × 1000(37-5)/0.4 w
={2.2 ×32/0.4 } w
=5.5 × 32 w =176 w
dQ/ dt = KAdT/dL
where
dQ/dt is rate of change of heat , K is thermal conductivity , A is cross-section area , dT is the temperature difference , and dL is thickness of wire .
now,
dQ/dt = 2.2 × 10^-3 × 1000(37-5)/0.4 w
={2.2 ×32/0.4 } w
=5.5 × 32 w =176 w
Answered by
5
Answer:
Explanation:176w
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