When `0.2 kg` of brass at `100 .^(@) C` is dropped into `0.5 kg` of water at `20 .^(@) C`,the resulting temperature is `23 .^(@) C`. The specific heat of brass is.

Options

A. `0.41 xx 10^3 Jkg^-1 .^(@) C^-1`.
B. `0.41 xx 10^2 Jkg^-1 .^(@) C^-1`.
C. `0.41 xx 10^4 Jkg^-1 .^(@) C^-1`.
D. `0.41 Jkg^-1 .^(@) C^-1`.

Answer

A

Explanation

Heat lost = Heat gained
`:. m_1s_1DeltaT_1 = m_2s_2DeltaT_2`
`:. s_1 = (m_2s_2DeltaT_2)/(m_1DeltaT_1)`
= `(0.5 xx 4.2 xx 10^3 xx 3)/(0.2 xx 77)Jkg^-1 .^(@) C^-1`
`:. s_1 = 0.41 xx 10^3 Jkg^-1 .^(@) C^-1`.