1. Thermo-physical properties 2. Radiation properties - nptel - Indian ...

Mechanical Measurements Prof S.P.Venkatesan
Indian Institute of Technology Madras
Example 4
In a calorimeter experiment for determination of specific heat of glass
beads, 0.15 kg of beads at 80°C are dropped in to a calorimeter
which is equivalent to 0.5 kg of water, at an initial temperature of
20°C. The maximum temperature of 22.7°C of the mixture occurs by
linear increase of the temperature in 10 s after the mixing starts. The
subsequent cooling shows that the temperature drops at 0.2°C/min
when the temperature is 22.7°C. Estimate the specific heat of glass
beads. Assume that the specific heat of water is 4.186 kJ/kg°C.
The given data is written down using the nomenclature introduced in
the text:
m= 0.015 kg, T = 20 ° C, T = 22.7°
C
3
1 2
T = 80 ° C, C = 0.5× 4.186 = 2.093 kJ / ° C
ΔT
0.2
tmax = 10 s, = 0.2 ° C/ min = = 0.0033 ° C/ s
Δt
60
The cooling constant may be calculated as
ΔT
C
t 2.093× 0.0033
K = Δ = = 0.0026 kW / ° C
( T −T ) ( 22.7 −20)
2 1
The heat loss during the mixing process is then approximated as
( T T ) t
( )
2 − 1 max 22.7 − 20 × 10
QL≈ K = 0.0026× = 0.035 kW
2 2
The specific heat of glass bead sample may now be estimated as
( )
mT ( T)
3 2
( )
( )
C T2 − T1 + QL
2.09× 22.7 − 201 + 0.035
c= = = 0.661 kJ / kg° C
− 0.15× 80 −22.7