Saved Bookmarks
| 1. |
A beaker contains 0.1 kg of water, initially at room temperature (20^(@)C). If the specific heat of water is 4.2 kJ/kg*""^(@)C and the latent heat of vaporization is 2,300kJ/kg, how much thermal energy would the water need to absorb to turn completely to steam? |
|
Answer» Solution :First we NEED to HEAT the water to the boiling point, `100^(@)C`, then we need to continue to add heat until all the liquid is vaporized. The AMOUNT of THERMAL energy required of the first step is `Q_(1)=m_("water")c_("water")(100^(@)C-20^(@)C)` `=(0.1kg)(4.2kJ//kg*""^(@)C)(80^(@)C)` `=33.6 kJ` and the amount of energy required for the second step is `Q_(2)=m_("water")L_("vap")=(0.1kg)(2,300kg//kg)=230kJ` Therefore, the total amount of thermal energy requires is `Q_("total")=Q_(1)+Q_(2)=33.6 kJ+230kJ=264kJ`. |
|