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Water has the specific heat 4,186kJ/kg*""^(@)C, a boiling point of 100^(@)C, and a heat of vaporisation of 2,260kJ/kg. a sealed beaker contains 100g of water that's initially at 20^(@)C. If the water absorbs 100 kJ of heat, what will its final temperature be? |
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Answer» `100^(@)C` `Q=mcDeltaT=(0.1kg)(4.186kJ//kg*""^(@)C)(100^(@)C-20^(@)C)=33kJ` Once the water REACHES `100^(@)C` any ADDITIONAL heat will be absorbed and begin the transformation to steam. to completely VAPORIZE the sample requires. `Q=mL=(0.1kg)(2260kJ//kg)=226kJ`. SINCE the `20^(@)C` water absorbed only 100 kJ of heat, enough heat was provided to bring the water to boiling, but enough to completely vaporize it (at which point, the absorption of more heat would begin to increase the temperature of the steam). thus, the water will reach and remain at `100^(@)C`. |
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