Question

Q2(a) A Pyrex beaker of 400 ml (coefficient of linear expansion = 0.3x10-5 °C-+) is filled with 350 ml of glycerine (coefficient of volume expansion = 5.1x10+ °C!). Analyze if the glycerine will overflow if the temperature of both beaker and glycerine is heated from 25 °C to 100 °C. (5 marks) (b) Figure Q2(b) shows the composite wall of a freezing plant which consists of a 10 cm layer of corkboard on the inside and 14 cm layer of concrete on the outside, with an area of 1 m². The temperature on the inside corkboard surface is -20 °C, and the temperature on the outside concrete surface is 24 °C. The thermal conductivity for corkboard is 0.04 W/m.K and for concrete is 0.8 W/m.K. -20°C Cork Concrete 1 = 24°C + 10 cm cm Figure Q2(b) Page 1 of 2 (i) Calculate the temperature of the interface between the two materials. (3 marks) Find the rate of heat loss for both the corkboard and concrete. (2 marks) (ii)

Answer 1

Volume expansion of Glass beaker is given as

V_g = V_go [ 1 + $\gamma$ _g$\Delta$T ] .......................... (1)

where V_g is volume of glass beaker after temperature increase $\Delta$ T = (100-25) = 75^o C .

V_go is initial volume and $\gamma$ _g = 0.3 $\times$ 10^-5oC^-1 is volume expansion coefficient of glass.

Expanded volume of glass beaker after temperature increase of 75^o C is calculated from equation (1) as

V_g = 400 [ 1 + 0.3 $\times$ 10^-5$\times$ 75 ] = 400.09 ml

Volume expansion of glycerine is given as

V_l = V_lo [ 1 + $\gamma$ _l$\Delta$T ] .......................... (2)

V_lo = Volume of glycerine = 350 ml and $\gamma$ _l = 5.1 $\times$ 10^-4oC^-1 is volume expansion coefficient of glycerine.

Expanded volume of glycerine liquid after temperature increase of 75^o C is calculated from equation (2) as

V_l = 350 [ 1 + 5.1 $\times$ 10^-4$\times$ 75 ] = 363.39 ml

Expanded volume of glycerine is less than the expanded volume of glass beaker. Hence glycerine will not over flow.

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ATA-KAT2 T= -20°C Cork Concrete T = 24°C 14 cm 10 cm

Let k₁ = 0.04 W m^-1 K^-1 be thermal conductivity of corck and k₂ = 0.8 W m^-1 K^-1

Under steady state , heat flux through cork and concrete will be equal

Hence we have ,

.........................(3)

Ti ka k2.45 ?Tܠ ?Tܠ 1ܠ

where A is common area of cork and concrete wall , $\Delta$ T₁ is the temperature difference between cork surface to interface, $\Delta$ x₁ is the depth of cork wall , $\Delta$ T₂ is the temperature difference between interface to concrete surface and $\Delta$ x₂ is the depth of concrete wall.

By substituting known values in eqn.(3) , we get

0.04 $\times$ ( $\Delta$ T₁/ 0.10 ) = 0.8 $\times$ ( $\Delta$ T₂ / 0.14 )

we get $\Delta$ T₂ = 0.07 $\Delta$ T₁   ........................(4)

Hence $\Delta$ T₁ + $\Delta$ T₂ = 1.07 $\Delta$ T₁ = 44 or   $\Delta$ T₁ = 41.12 ^oC

Hence Interface temperature = -20 + 41.12 = 21.12 ^oC

Heat loss = 0.04 $\times$ ( 41.12/ 0.10 ) = 16.45 W = 16.45 J/s