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
0 votes
642 views
in Physics by (32.9k points)
closed by

A blackbody with initial temperature of 300 °C is allowed to cool inside an evacuated enclosure surrounded by melting ice at the rate of 0.35 °C/second. If the mass, specific heat and surface area of the body are 32 grams, 0.10 cal/g °C and 8 cm2 respectively, calculate Stefan’s constant. (Take J = 4200 j/kcal.)

1 Answer

+1 vote
by (32.7k points)
selected by
 
Best answer

Data : T = 273 + 300 = 573 K, T0 = 273 K,

\(\frac{dQ}{dt}\) = 0.35 °C/s = 0.35 K/s, at

M = 32 g = 32 × 10-3 kg, A = 8 cm2 = 8 × 10-4

m2C = 0.10 cal/g.°C = 0.10 kcal/kg.K = 420 j/kg.K

since J = 4200 J/kcal

Stefan's constant, σ = 5.754 x 10-8 W/m2.k4

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.

Categories

...