Question

A body cools from 80⁰C to 50⁰C in 5 minute. Calculate the time it takes to cool from 60⁰C to 30⁰C. The temperature of surroundings is20⁰C –

• A. 6 min
• B. 4.5 min
• C. 9 min
• D. 12 min

Answer 1

Answer: (C)

9 min

Answer 2

In first case:

Average temperature of liquid

=$\frac{80 + 50}{2}$= 65⁰C

Excess temp = (65 –20) ⁰C = 45⁰C

$\frac{d\theta_{1}}{dt}$=$\frac{50 - 80}{5}$= – 6⁰C/min.

– 6 = K × 45 … (1)

In second case:

Average temperature of liquid =$\frac{60 + 30}{2}$= 45⁰C.

Excess temp = (45 – 20) ⁰C = 25⁰.

Rate of fall of temp $\frac{d\theta_{2}}{dt}$

$\frac{d\theta_{2}}{dt}$= –$\frac{60 - 30}{t_{\min}}$

–$\frac{30}{t_{\min}}$= K × 25 … (2)

Divide (1) by (2) t = 9 min.

$\frac{- 6}{\frac{- 30}{t_{\min}}}$=$\frac{K \times 45}{K \times 25}$

t_min =$\frac{45}{25}$×$\frac{5}{1}$= 9 min.