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Question
Solve the following problem.
Find the temperature difference between two sides of a steel plate 4 cm thick, when the heat is transmitted through the plate at the rate of 400 k cal per minute per square meter at a steady state. Thermal conductivity of steel is 0.026 kcal/m s K.
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Solution
Given: `"Q"/"A t"` = 400 kcal/min m2 = `400/60` kcal/s m2
x = 4 cm = 4 × 10-2 m, k = 0.026 kcal/m s k
To find: Temperature difference (T1 - T2).
Formula: Q = `("kA"("T"_1 - "T"_2))/"x"`
Calculation: From formula,
`"T"_1 - "T"_2 = "Q"/"At" xx "x"/"k"`
∴ `"T"_1 - "T"_2 = 400/60 xx (4 xx 10^-2)/0.026 = 10.26`K
The temperature difference between the two sides is 10.26 K.
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