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A slab of stone of area 0.36m2and thickness 0.1is exposed on the lower surface to steam at 100C. A block of ice rests 0C rests on the upper surface of the slab. In one hour 4.8kg of ice melted. The thermal conductivity of the slab is: (Given latent heat of fusion of ice 3.36×105Jkg1)


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Solution

Step. 1 Given Data,

Latent heat of fusion of ice, L=3.36×105Jkg1

m is the mass of the given sample=4.8kg
x is the thickness of the slab=0.1m
Ais the area=0.36m2

A slab of stone is exposed on the lower surface to steam at T1,100C

A block of ice rests T2=0C rests on the upper surface of the slab
dT is the time given=1hour=3600sec

Step 2. Formula used,
Lfis the latent heat of fusion.
dQ=mLf
dQis the change in heat

m is the mass of the given sample

dQdT=KA(T1T2)x
K is the thermal conductivity of slab

A is the area

dT is the change of temperature

x is the thickness

Step 3. Calculating the thermal conductivity
Fourier’s Law of heat conduction is
dQdT=KA(T1T2)x......(i)
dT is the time given=1hour=3600sec
A is the area=0.36m2
x is the thickness of the slab=0.1m
T1T2 is the difference in temperature(1000)=100
Substituting all the values in equation (i),
mLfdT=KA(T1T2)x
K=mLfxdT.A.(T1T2)
K=4.8×3.36×105×0.13600×0.36×100
Solving the equation,
K=4.8×3.360.36×36
K=1.24J/m/s/C
The thermal conductivity of the slab is 1.24J/m/s/C.


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