question_answer 19)

A cubical ice box of thermo Cole has each side = 30 cm and a thickness of 5 cm. 4 kg of ice is put in the box. If outside temperature is \[{{45}^{o}}C\] and coeff. of thermal conductivity \[\frac{1\cdot 05-x}{x}=\frac{{{\text{F}}_{\text{1}}}}{{{\text{F}}_{\text{2}}}}\text{=}\frac{\text{1}}{\text{2}}\] calculate the mass of ice left after 6 hours. Take latent heat of fusion of ice \[2\cdot 10-2x=x\]

Answer:

Here, length of each side, \[x=0\cdot \text{70 m=70 cm}\] Thickness of each side \[{{\text{F}}_{\text{1}}}\text{ and }{{\text{F}}_{\text{2}}}\] Total surface area through which heat enters into the box, \[\frac{{{\text{F}}_{\text{1}}}}{{{\text{A}}_{\text{1}}}{{\text{Y}}_{\text{1}}}}\text{=}\frac{{{\text{F}}_{\text{2}}}}{{{\text{A}}_{\text{2}}}{{\text{Y}}_{\text{2}}}}\] Temp. diff., \[\frac{{{\text{F}}_{\text{1}}}}{{{\text{F}}_{\text{2}}}}\text{=}\frac{{{\text{A}}_{\text{1}}}{{\text{Y}}_{\text{2}}}}{{{\text{A}}_{\text{2}}}{{\text{Y}}_{\text{2}}}}\text{=}\]\[\frac{\text{1}}{\text{2}}\text{ }\!\!\times\!\!\text{ }\frac{\text{2 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{11}}}}{\text{7 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{10}}}}\text{=}\frac{\text{10}}{\text{7}}\]\[{{F}_{1}}x={{F}_{2}}\left( 1\cdot 05-x \right)\] time, \[\frac{1\cdot 05-x}{x}=\frac{{{\text{F}}_{\text{1}}}}{{{\text{F}}_{\text{2}}}}\text{=}\frac{\text{10}}{\text{7}}\] Latent heat of fusion, \[10x=7\cdot 35-7x\] Let m be the mass of ice melted in this time \[x=0\cdot 4324m\] \[=43\cdot \text{2 cm}\] \[1\cdot 0\] \[0\cdot 5\times {{10}^{-2}}c{{m}^{2}}\] \[g=10m{{s}^{-2}};\] \[Y=2\times {{10}^{11}}N{{m}^{-2}}.\]Mass of ice left \[AC=CB=l=0\cdot 5m;\]