Question

Emily is testing her baby's bath water and finds that it is too cold, so she adds some hot water from a kettle on the stove. If Emily adds 3.00 kg of water at 79.0 °C to 27.0 kg of bath water at 25.0 °C, what is the final temperature of the bath water?

Answer 1

The final temperature of the bath water is
`30.4^(\circ)`

Solution:

From given,

Mass of the hot water = m = 3 kg

Mass of cold water = M = 27 kg

Initial temperature of the hot water =
`T_(1h) = 79^(\circ)`

Initial temperature of the cold water =
`T_(1c) = 25^(\circ)`

The heat lost by the hot water will be equal to the heat gain by the cold water

Therefore,

`mC(T_(1h) - T) = MC(T - T_(1c))`

`3 * 4.18 (79-T) = 27 * 4.18(T - 25)\\\\12.54(79-T) = 112.86(T - 25)\\\\990.66 - 12.54T = 112.86T - 2821.5\\\\112.86T + 12.54T = 990.66 + 2821.5\\\\125.4T = 3812.16\\\\Divide\ both\ sides\ by\ 125.4\\\\T = 30.4`

Thus the final temperature of the bath water is
`30.4^(\circ)`