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A large container is separated into two halves by a 2.0 cm thick glass divider. The area of the divider is 0.25 m^2 and the the...Asked by sand
A large container is separated into two halves by a 2.0 cm thick glass divider. The area of the divider is 0.25 m^2 and the thermal conductivity of glass is 0.90W/m•K. Each side is filled with 30 kg of water, specific heat 4180 J/kg•K, at different temperatures. Initially, the temperature difference is 20 K. How long will it take before the temperature difference is 1.0 K?
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Answered by
drwls
Assume the water on each side of the divider is at uniform but changing temperature, due to convection or stirring.
Let the two temperatures on opposite sides be T1 and T2
The rate of heat conduction through the divider is
dQ/dt = (T1- T2)*k*A/(thickness)
= (T1 - T2)(0.9)(0.25)/0.02
= (T1 - T2)*11.25 J/(K*s)
The energy conservation equation is
dT1/dt = -dT2/dt = -(1/MC)dQ/dt
d/dt(T1 - T2) = -[2/(M*C])*dQ/dt
= -1.595*10^-5 (K/J) dQ/dt
Let X be the temperature difference, T1 - T2
dX/dt = -1.595*10^-5 dQ/dt
= -1.79*10^-4 X
dX/X = -1.794*10^-4 *t
X = Xo*exp(-t/5573 s)
Where Xo = 20K is the inityial tyemperature difference.
Solve for t when X = 1 C
Let the two temperatures on opposite sides be T1 and T2
The rate of heat conduction through the divider is
dQ/dt = (T1- T2)*k*A/(thickness)
= (T1 - T2)(0.9)(0.25)/0.02
= (T1 - T2)*11.25 J/(K*s)
The energy conservation equation is
dT1/dt = -dT2/dt = -(1/MC)dQ/dt
d/dt(T1 - T2) = -[2/(M*C])*dQ/dt
= -1.595*10^-5 (K/J) dQ/dt
Let X be the temperature difference, T1 - T2
dX/dt = -1.595*10^-5 dQ/dt
= -1.79*10^-4 X
dX/X = -1.794*10^-4 *t
X = Xo*exp(-t/5573 s)
Where Xo = 20K is the inityial tyemperature difference.
Solve for t when X = 1 C
Answered by
sand
Thanks a lot.
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