A coil of enamelled copper wire of resistance `50 Omega` is embedded in a block of ice and a potential difference of 210 V applied across it. Calculate, how much ice will melt in half minute. Latent heat of ice is 180 cal per gram.

A coil of enamelled copper wire of resistance `50 Omega` is embedded i

1.	A coil of enamelled copper wire of resistance `50 Omega` is embedded in a block of ice and a potential difference of 210 V applied across it. Calculate, how much ice will melt in half minute. Latent heat of ice is 180 cal per gram.
Answer» Here, `R = 50 Omega, V = 210, t = 30s, L = 80 cal//g` Heat produced, `H = (V^(2)t)/4.2R = ((210)^(2)xx30)/(4.2 xx 50) = 210 xx 30 cal` Let m gram of ice be melted in half minute, then `mL = H or m = H/L = (210 xx 30)/80 = 78.75g`

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A coil of enamelled copper wire of resistance `50 Omega` is embedded in a block of ice and a potential difference of 210 V applied across it. Calculate, how much ice will melt in half minute. Latent heat of ice is 180 cal per gram.

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