An ice cube whose mass is `50 g` is taken from a refrigerator where its temperature was `-10^@C`. If no heat is gained or lost from outside, how much water will freeze onto the cube if it is dropped into a beaker containing water at `0^@C`? Latent heat of fusion`=80 kcal//kg`, specific heat capacity of ice`=500 cal//kg//K`.

Options

A. `2.265g`
B. `3.125g`
C. `5.425g`
D. `4.355g`

Answer

B

Explanation

Let m kg of water freeze on it.
Then heat lost`=mxx80xx1000cal`
Heat gained by ice`=50xx10^-3xx500xx{0-(-10)}cal` Now, Heat loss`=` Heat gain, `m=3.125xx10^-3kg=3.125g`