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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.
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.
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.
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 ---- Please feel free to post as many doubts on our disucssion forum as you can. If you find any question difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation. All the best. Regards, Naga Ramesh IIT Kgp - 2005 batch
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
----
Please feel free to post as many doubts on our disucssion forum as you can. If you find any question difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation. All the best. Regards, Naga Ramesh IIT Kgp - 2005 batch
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
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
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