Question

Ch18.40

Calculate the specific heat of a metal from the following data. A container made of the metal has a mass of 3.6 kg and contains 17 kg of water. A 2.0 kg piece of the metal initially at a temperature of 194

Answer 1

Mass of Metal 'M' = 3.6 kg= 3600g
Mass of water 'm' = 17 kg= 17000g

Mass of metal dropped into water m' = 2kg= 2000g and its temperature T' = 194C
Initial Temperature of M and m = 16C
Final temperature of M,m and m' = 18C

Since,
Q=mc(Delta)T
Q-> Heat absored/released
m -> mass
c-> heat capacity of metal)
(Delta)T -> Change in temperature

Situation I: Heat released by metal at temperature of 194C
Q = 2000 x c x (194-18)= 352,000c

Situation II: Heat absrbed by Metal (M) and mass system (m')
Q= (M).(c_M).(Delta)T + m.(c_H2o).(Delta)T= 3,600c(18-16) + 17,000x4.186x(18-16) = 7,200c+ 142,324

Assuming no energy is lost
Situation I = Situation II

352,000c = 7,200c+ 142,324
344800c= 142,324
c = 0.413 Jg^-1C^-1