Specific heat of copper 420JKg−1K−1
Heat energy absorbed:
Heat energy absorbed:
Q=mcΔT....(1)
Work done is equal to the loss in mechanical
energy
ΔE=W
=(mgsinθ)L........(2)
From (1) and (2)
mcΔT=mgsinθL
ΔT=gsinθLc
ΔT=10sin(37)×0.6420
ΔT=8.5×10−30C