A) \[150{}^\circ C\]
B) \[110{}^\circ C\]
C) \[130{}^\circ C\]
D) \[200{}^\circ C\]
Correct Answer: B
Solution :
| suppose area of the bottom of the tank =\[A\,c{{m}^{3}}\] |
| Volume of water that vaporises in 9 min \[\left( or\,540s \right)\] \[=A\times 1c{{m}^{3}}\] |
| Mass of water that vaporises in 540s\[=A\,c{{m}^{3}}\times 1gc{{m}^{-3}}=Ag\] |
| \[\therefore Q=mL=A\times 540cal\] |
| But \[Q=\frac{KA({{T}_{1}}-{{T}_{2}})}{x}\times t\] Or \[{{T}_{1}}-{{T}_{2}}=\frac{Qx}{KAt}\] \[=\frac{A\times 540\times 2}{0.2\times A\times 540}=10\] |
| Total temperature of the furnace i.e., \[{{T}_{1}}={{T}_{2}}+10=100+10=110{}^\circ C\] |
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