A metal block of heat capacity 90 J/°C placed in a room at 25°C is heated electrically. The heater is switched off when the temperature reaches 35°C. The temperature of the block rises at the rate of 2°C/s just after the heater is switched on and falls at the rate of 0.2 °C/s just after the heater is switched off. Assume Newton’s law of cooling to hold.
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
[ 180 W 18 W
9 W 
s]
t = 
= 5 sec.
P × t = 90 ×10
or P × 5 = 90 × 10
P = 180 W
P' = 
= 90 × 0.2 = 18 W
or P 30 = 9W ( P 35 = 18 W)
P × t = C 
T + Q lost
180 × t = 90 × 10 + 9 × t
or t = 
= 
sec.
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems