A coil of wire of resistance 50 ohm is embedded in a block of ice and a potential difference of 210 volt is applied across it the amount of ice which melt in one second is?
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Answer:
pls see the attachment
Given,
resistance of the coil of wire , R = 50 Ω
potential difference, V = 210 Volts.
To find,
the amount of ice which melted in one second.
energy dissipation due to electric current passing through the coil of wire, E = V²t/R
= (210)²t/50
= 882t Joule
we know, latent heat of ice, Lf = 336 × 10³ J/Kg
from Calorimetry,
heat lost due to electric current = heat gained by ice to melt it.
⇒882t J = m × 336 × 10³ J/kg
⇒m/t = 882/(336 × 10³) kg = 2.625 × 10¯³ kg
⇒m/t = mass of ice melted er second = 2.625 g
Therefore 2.625 g of ice which melted in one second.
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