Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
89 views
in Physics by (89.9k points)
closed by
Water is being boiled in flat bottom kettle placed on a stove. The area of the bottom is `3000cm^2` and the thickness is 2 mm. If the amount of steam produced is `1g//min`, calculate the difference of temperature between the inner and outer surface of the bottom. K for the material of kettle is 0.5`cal^@C//s//cm`, and the latent heat of steam is `540 cal//g`.
A. `0.12 xx 10^(-5)`K
B. `1.9 xx 10^(-3)`K
C. `1.3 xx 10^(-4^(@))`C
D. `1.2 xx 10^(-3)`K

1 Answer

0 votes
by (90.8k points)
selected by
 
Best answer
Correct Answer - D
d) Rate of production of mass of steam = `(dm)/(dt) = 1/(60) gs^(-1)`
Rate of heat transformed by stove,
`(dtheta)/(dt) = L(dm)/(dt)=540 xx 1/(60) = 9cal^(@)C^(-1)s^(-1)cm^(-1)`
Also, `(d(theta))/(dt) = (KAd(theta))/(dx)rArr L(dm)/(dt)=(KAd(theta))/dx`
`rArr 9 = (0.5 xx 3000xx d(theta))/(0.2)`
`rArr d(theta) = 1.2 xx 10^(-3^(@))C=1.2xx10^(-3)`K

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

...