A piece of metal with a mass of 97.6 grams was heated in a steam jacket to 94°C and then
plunged into 243.6 grams of water at 21 degrees * C causing the temperature to rise to 24 degrees * C What
is the specific heat of metal?
H=mcΔθH =mc\Delta\thetaH=mcΔθ
Heat lost by metal
=97.6×c×(94−24)=6832c= 97.6×c × (94-24)= 6832c=97.6×c×(94−24)=6832c
Heat gained by water
243.6×4.184×(24−21)=3057.7243.6 ×4.184×(24-21)= 3057.7243.6×4.184×(24−21)=3057.7
6832c=3057.7x=0.4476Jg−1K−16832c=3057.7\\x= 0.4476Jg^{-1}K^{-1}6832c=3057.7x=0.4476Jg−1K−1
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