Saved Bookmarks
| 1. |
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.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» Correct Answer - A 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`. |
|