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Question
Solve the following problem.
A metal sphere cools at the rate of 0.05 ºC/s when its temperature is 70 ºC and at the rate of 0.025 ºC/s when its temperature is 50 ºC. Determine the temperature of the surroundings and find the rate of cooling when the temperature of the metal sphere is 40 ºC.
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Solution
Given: T1 = 70 ºC, `("dT"/"dt")_1,` 0.05 ºC/s
T2 = 50 ºC, `("dT"/"dt")_2,` 0.025 ºC/s, T3 = 40 ºC.
To find: i. Temperature of surrounding (T0)
ii. Rate of cooling `("dT"/"dt")_3`
Formula: `"dT"/"dt" = "C"("T" - "T"_0)`
Calculation: From formula,
`("dT"/"dt")_1 = "C"("T"_1 - "T"_0) and ("dT"/"dt")_2 = "C"("T"_2 - "T"_0)`
∴ `0.05/0.025 = ("C"(70 - "T"_0))/("C"(50 - "T"_0))`
∴ 2 (50 - T0) = 70 - T0
∴ T0 = 30 πC
Substituting value of To,
0.05 = C(70 - 30)
∴ C = `0.05/40 = 0.00125`/s.
For T3 = 40 ºC
`("dT"/"dt")_3 = "C"("T"_3 - "T"_0) = 0.00125 (40 - 30)`
= 0.00125 × 10
= 0.0125 ºC/s.
- Temperature of surroundings is 30 ºC.
- Rate of cooling at 40 ºC is 0.0125 ºC/s.
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