A pot with a steel bottom 1-2 cm thick rests on a hot stove- The area of the bottom of the pot is 0-150 m2- The water inside the pot is at 100- C and 0-440 kg are evaporated every 5-0 minute- Find the temperature of the lower surface of the pot- which is in contact with the stove- Take Lv - 2-256 - 106 J-kg and ksteel - 50-2 W-m - K-in C

The heat flow from stove is used to convert water to steam. Hence, dQ dt = 0.44(2.256X 10 ^ 6 ) 5(60) =3309J/s Also, dQ dt = KA L Δ T= 50.2(0.15 ...

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Standard XII

Chemistry

Relative Lowering of Vapour Pressure

A pot with a ...

Question

A pot with a steel bottom 1.2 cm thick rests on a hot stove. The area of the bottom of the pot is $0.150 m^{2}$ . The water inside the pot is at $100^{\circ} C$ and 0.440 kg are evaporated every 5.0 minute. Find the temperature of the lower surface of the pot, which is in contact with the stove. Take $L_{v} = 2.256 \times 10^{6} J / k g$ and $k_{s t e e l} = 50.2 W / m - K$ .(in C)

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Solution

The heat flow from stove is used to convert water to steam. Hence,

$\frac{d Q}{d t} = \frac{0.44 (2.256 X 10^{6})}{5 (60)} = 3309 J / s$

Also, $\frac{d Q}{d t} = \frac{K A}{L} Δ T = \frac{50.2 (0.15)}{0.012} (T - 100)$

$627.5 (T - 100) = 3309$

$T = 105 {.27}^{0} C$

Answer is $105 {.27}^{0} C$

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The heat flow from stove is used to convert water to steam. Hence,dQdt=0.44(2.256X106)5(60)=3309J/sAlso, dQdt=KALΔT=50.2(0.15)0.012(T−100)627.5(T−100)=3309T=105.270CAnswer is 105.270C

The heat flow from stove is used to convert water to steam. Hence,

$\frac{d Q}{d t} = \frac{0.44 (2.256 X 10^{6})}{5 (60)} = 3309 J / s$

Also, $\frac{d Q}{d t} = \frac{K A}{L} Δ T = \frac{50.2 (0.15)}{0.012} (T - 100)$

$627.5 (T - 100) = 3309$

$T = 105 {.27}^{0} C$

Answer is $105 {.27}^{0} C$