Heating Mineral Water to 60°C Using Plain Water at 100°C

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To heat 1 kg of mineral water from 0°C to at least 60°C using 1 kg of plain water at 100°C, the heat gained by the mineral water must equal the heat lost by the plain water. The equation used is ΔQ = mcΔT, where specific heat capacities (cp) for the two types of water may differ due to temperature variations. The calculations indicate that more than 1 kg of plain water would be required, suggesting a potential oversight in the specific heat values. The discussion highlights the need to consider the differences in specific heat capacities for accurate results. Further exploration of the problem is necessary to identify the correct approach.
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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.
 
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Check your cp's. They should be different given the temperature difference and though subtle, material difference.
 
Thanks but don't think that there are so much difference between them. Any other ideas?
 
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