500 kg of water is heated from `20^(@)` to `100^(@)` C . Calculate the increase in the mass of water.Given specific heat of water `=4.2 xx 10^(3) J kg ^(-1) .^(@)C^(-1)`.

Options

A. `3.2 xx 10^(-9)`kg
B. `1.87 xx 10^(-9)`kg
C. `0.96 xx 10^(-9)`kg
D. `2.8 xx 10^(-9)` kg

Answer

B

Explanation

Given, m = 500 kg
Rise in temperature, `dT=(100^(@)-20^(@)C = 80^(@)`C
Specific heat of water, c`=10^(3)` cal `kg^(-1) C^(-1)`
Heat energy gained by water, E=mcdT
`=500 xx 10^(3) xx 80`cal = `4 xx 10^(7) xx 4.2`J = `16.8 xx 10^(7)`J
From , `E=mc^(2)`
Increase in mass of water,
`m=E/c^(2) = (16.8 xx 10^(7))/(3 xx 10^(8))^(2) = 1.87 xx 10^(-9)` kg