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Homework Statement
How can you heat 1 \,\mathrm{kg} mineral water which is at a temperature of 0 \,{}^\circ \mathrm{C} to at least 60 \,{}^\circ \mathrm{C} with using 1 \,\mathrm{kg} plain water at a temperature of 100 \,{}^\circ \mathrm{C}?
Homework Equations
\Delta Q = mc\Delta T
The Attempt at a Solution
\underbrace{(1000 g)(4.18~J/g~C)(60~C)}_{\text{heat gained}} = \underbrace{m(4.18~J/g~C)(100~C - 60~C)}_{\text{heat lost}}
But then m>1000 g. So I really don't know how to do it. There should be a nasty trick in it.