A coil of wire of resistance `50 Omega` is embedded in a block of ice and a potential difference of 210 V is applied across it. The amount of ice which melts in 1 second is [ latent heat of fusion of ice `= 80 cal g^(-1)`]

Options

A. 0.262 g
B. 2.62 g
C. 26.2 g
D. 0.0262 g

Answer

B

Explanation

Heat energy of melting is provided by electric energy. Therefore, `mL = V^(2)/R t`, where m, L, V, R & t represents mass of ice, latent heat of ice, voltage across resistance, resistance and time respectively.
On substituting all the corresponding values in the above relation, we have
`rArr m(80) xx 4.2 = (210)^(2)/50 (1)`
`rArr m = 2.652 g`