24000J heat is applied to increase the temperature of a rod of length
1 m having mass 3 kg from , whose expansion of
Length of another similar rod is 2.2× m for the same change of
temperature.
1)Find the specific heat of the 1st rod.
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Explanation:
Free expansion of the rod =αLΔθ
=15×10
−60
C×2m×(50−20)
0
C
=9×10
−
4m=0.9mm
If the expansion is fully prevented, then
Strain=
2
9×10
−4
=4.5×10
−4
Temperature Stress = Strain ×Y
=4.5×10
−4
×2×10
11
=9×10
7
N/m
2
If 0.4 mm expansion is allowed, then length restricted to expand
=0.9−0.4=0.5mm
Strain=
2
5×10
−4
=2.5×10
−4
Temperature stress = Strain ×Y=2.5×10
−4
×2×10
11
=5×10
7
N/m
2
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