A load of 31.4 kg is suspended from a wire of radius 10-3 m and density 9*103 kg.m-3 . Calculate the change in temperature of the wire of the wire if 75 % of the work done is converted into heat. The Young's modulus and specific heat capacity of the material of the wire are 9.8 * 1010 N.m-2 and 490 J.kg-1.K-1 respectively.
[ Answer : 0.0083 oC ]
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Answers
Change in temperature of the wire of the wire if 75 % of the work done is converted into heat is :
• Given : mass of load, m = 31.4 kg, density, p = 9×10^3 kg/m^3, radius of wire = 10-3 m
Young modulus = 9.8×10^10 N/m
Specific heat = 490 J/kg K
•Also, 75% of energy in converted to heat. i.e Q = 75% of U
Volume of wire = pi×r^2×L
mass of wire = pi×r^2×L×p
• Let ∆L be the increase in the length of wire by a load of mass M then,
• Stress = F/A = Mg / pi×r^2
• Strain = ∆L / L
• We know that,
Potential energy per unit volume is given as
u = 1/2×stress×strain
= Mg×∆L / 2×pi×r^2×L
• Elastic potential energy = u×volume
= u×pi×r^2L = 1/2×Mg∆L .... (1)
• Young modulus, Y = Stress / strain
= (Mg/pi×r^2)×(L/∆L)
∆L = MgL/pi×r^2Y
Substituting ∆L = MgL/pi×r^2Y in (1)
• U = 1/2×(Mg)^2L/pi×r^2Y
• We already know that,
Q =75/100×U
= 75/100×1/2×(Mg)^2L/pi×r^2Y
• If ∆T is temperature rise then,
Q = ms∆T
∆T = Q/ms
={75/100×1/2×(Mg)^2L/pi×r^2Y}×1/ms
• Substitute all given values in above equation, we'll get
• ∆T = 0.0083°C