Use app×
QUIZARD
QUIZARD
JEE MAIN 2026 Crash Course
NEET 2026 Crash Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
+1 vote
1.4k views
in Physics by (51.2k points)
closed by

Temperature of hot soup in a bowl goes 98°C to 86°C in 2 minutes. The temperature of surrounding is 22°C. Find the time taken for the temperature of soup to go from 75°C to 69°C. (Assume newton's law of cooling is valid)

(1) 4.4 minute.
(2) 3.6 minute.
(3) 6.2 minute.
(4) 1.4 minute.

1 Answer

+2 votes
by (50.5k points)
selected by
 
Best answer

(4) 1.4 minute

Using Newton's law of cooling, Rate of change in temperature, R = dT/dt = -k(T - Ts), T = \(\frac{T_i+T_f}{2}\) 

Divide (i) & (ii)

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

...