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A coil of wire of resistance 50 Omega is...

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)`]

A

0.262 g

B

2.62 g

C

26.2 g

D

0.0262 g

Text Solution

Verified by Experts

The correct Answer is:
B
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`
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Knowledge Check

  • A coil of water 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) ]

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

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    A
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    B
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    C
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    D
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    C
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    D
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