wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Heat is being supplied at a constant rate to a sphere of ice which is melting at the rate of 0.1 gm/sec. It melts completely in 100 sec. The rate of rise of temperature thereafter will be (Assume no loss of heat)

A
0.8 C/sec
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
5.4 C/sec
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
3.6 C/sec
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
Will change with time
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 0.8 C/sec
Required heat to melt 0.1 grams of ice per second is ΔQ/Δt=0.1×80=8 cal/s
Amount of ice converted to water is =0.1×100=10 gm

Rate of rise in temperature will be ΔTΔt=ΔQ/Δtms=8/(10×1)=0.8 C/s.

flag
Suggest Corrections
thumbs-up
57
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Thermal Expansion
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon