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If you have 430.0 mL of water at 25.00 °C and add 140.0 mL of water at 95.00 °C, what is the final temperature of the mixture? Use 1.00 g/mL as the density of water.

Respuesta :

Edufirst
Heat released by the hot water = mCeΔT = 140.0 g Ce ( 95.00°C - T)

Heat absorbed by the cold water = mCeΔT = 430.0 g Ce (T - 25.00°C)

Heat released = heat absorbed

140.0*Ce(95 - T) = 430.0*Ce(T -25)

95 - T = 3.0714 (T - 25)

95 - T = 3.0717T - 76.786

3.071T + T = 95 + 76.786

4.071T =   171.786

T = 42.20 °C

Answer: 42.20 °C

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