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`        Ques) A coil of current crrying Nichrome wire is immersed in aliquid. When the potetial difference across the coil is 12 V and the current through the coil is 5.2 A., the liquidm evaporated at the steady rate of 21mg/sec. Calculate the heat of  Vapourization of the liquid.`
8 years ago

18 Points
```										Power consumed = V*I = 12 *5.2 = 62.4 J/s.
Rate of evapoation = 21 mg/s = 2.1 * 10-5  kg /s
Power consumed = Heat gaines by water per second
Heat of Vaporization, H  = 62.4 / (2.1 * 10-5)  =  2971 KJ/kg =710 Kcal / kg.
```
8 years ago
Ramesh V
70 Points
```										heat generated in coil (H) = I2Rt
given V =12 and I = 5.2 A , by ohms law, R = V/I = 12/5.2 =2.3 ohms
dH/dt = I2R  = 62.2 J/sec
let specific heat of vapourisation is:  L J/g
rate of evaporation is : m' = 21x10-3 g/sec
so, dH/dt = m' * L
L = (dH/dt)/ m'
L= 62.2 / 21x10-3
L = 2961.5 J/gram
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```
8 years ago
Vanya Saxena
18 Points
```										Heat generated per sec due to the current carrying coil,
H=VI  =>H =12*5.2 J/sec
This heat will be equal to the latent heat of vapourization as the same is responsible for the vapourization of the liquid,
VI=mL      =>12*5.2=21/1000*L
L=29.71*10^2 J/Kg

```
8 years ago
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