Question

A liquid takes 5 minutes to cool from 80^@C to 50^@C . How much time will it take to cool from 60^@C to 30^@C? The temperature of surroundings is 20^@C.

Answer 1

Time taken by the liquid to cool from 60°C to 30°C = 5 minutes.

• A liquid takes 5 minutes to cool from 80°C to 50°C. Time taken to cool from 60°C to 30°C. Surrounding temperature is 20°C.
• Rate of cooling = (initial temperature-final temperature)/(time taken to achieve temperature difference)
• Time taken to cool t = Temperature difference/R
• In case 1; T(initial)=80°C  T(final)=50°C t=5 min
• Rate of cooling R = (80-50)/5 = 6 °C/min
• In case 2; T(initial)=60°C  T(final)=30°C
• Time taken t=(60-30)/6=5 min.