Sum

**Solve the following problem.**

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?

Advertisement Remove all ads

#### Solution

**Given:** T_{1} = 80 °C, T_{2} = 60 °C, T_{3} = 40 °C, T_{0} = 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//15)` = 10 min

Time taken in cooling is **10 min**.

Concept: Newton’s Laws of Cooling

Is there an error in this question or solution?

Advertisement Remove all ads

#### APPEARS IN

Advertisement Remove all ads

Advertisement Remove all ads