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A metal sphere cools from 80 °C to 60 °C in 6 min. How much time with it take to cool from 60 °C to 40 °C if the room temperature is 30 °C?

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Question

A metal sphere cools from 80 °C to 60 °C in 6 min. How much time with it take to cool from 60 °C to 40 °C if the room temperature is 30 °C?

Numerical
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Solution

Given: T1 = 80 °C, T2 = 60 °C, T3 = 40 °C, T0 = 30 °C, (dt)1 = 6 min.

To find: Time taken in cooling (dt)2

Formula: `"dT"/"dt" = "C"("T" - "T"_0)`

Calculation: From formula,

`("dT"/"dt")_1 = "C"("T"_1 - "T"_0)`

∴ `(80 - 60)/6` = C(80 - 30)

∴ C = `20/(6 xx 50) = 1/15`/min

Now, `("dT"/"dt")_2 = "C"("T"_2 - "T"_0)`

∴ `(60 - 40)/("dt")_2 = 1/15 (60 - 30)`

∴ `"dt"_2 = (60 - 40)/30 xx 15`

∴ `"dt"_2 = 20/30 xx 15`

∴ `"dt"_2 = 300/30`

∴ `"dt"_2` = 10 min

Time taken in cooling is 10 min.

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Chapter 7: Thermal Properties of Matter - Exercises [Page 141]

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Balbharati Physics [English] Standard 11 Maharashtra State Board
Chapter 7 Thermal Properties of Matter
Exercises | Q 3. (xv) | Page 141

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