A bowl filled with very hot soup cools from 98 ° C to $86^{\circ} \mathrm{C}$ in 2 minutes when the…

A bowl filled with very hot soup cools from 98°C to $86^{\circ} \mathrm{C}$ in 2 minutes when the room temperature is $22^{\circ} \mathrm{C}$. How long it will take to cool from $75^{\circ} \mathrm{C}$ to $69^{\circ} \mathrm{C}$ ?
  1. 2 minute

  2. 1.4 minute
  3. 0.5 minute
  4. 1 minute

Solution

From Newton's law of cooling, we have

ΔQΔt=-KTavg-T0

Where, Q=msTm is the mass, s is the specific heat and T is the temperature change.

Case 1:

t=2 minutes, T0=22°C

ms×122=-K98+862-22

6=-Kms70    ...1

Case 2:

ms×6Δt=-K75+692-22

6Δt=-Kms(50)   ...2

Dividing (2) by (1), we get

6Δt(6)=5070

Δt=75=1.4 min

Asked in: JEE Main 2023 (25 Jan Shift 1)

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