Q. 6

# A cube of iron (d

Answer :

Given:

iron (density = 8000 kg m–3, specific heat capacity = 470 J kg–1 K–1)

density of ice = 900 kg m–3

latent heat of fusion of ice = 3.36 × 105 J kg–1.

Formula used:

ΔQ=MCΔT

ΔQ=heat exchange

ΔT=tempeature change

M=mass

C=heat capacity

Heat lost by iron= Heat gained by ice

8000× Volume× 470× T=900× Volume × 3.36× 105

T=80.42° C

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses
RELATED QUESTIONS :

A ball is droppedHC Verma - Concepts of Physics Part 2

A van of mass 150HC Verma - Concepts of Physics Part 2

The temperature oHC Verma - Concepts of Physics Part 2

A 50 kg man is ruHC Verma - Concepts of Physics Part 2

Calculate the timHC Verma - Concepts of Physics Part 2

Indian style of cHC Verma - Concepts of Physics Part 2

On a winter day tHC Verma - Concepts of Physics Part 2

The temperature oHC Verma - Concepts of Physics Part 2

If heat is suppliHC Verma - Concepts of Physics Part 2

When a solid meltHC Verma - Concepts of Physics Part 2