100 grams of water at 40 degree celcius is added to 50 grams of water at 60 degree celcius. find the temperature of the mixture?
Answers
Specific Heat of water is defined as the amount of heat (added or subtracted) in order to change the temperature of 1 gram of water by 1 degrees Kelvin (or Celsius), under constant pressure.
Let's assume a constant value (c) for the Specific Heat of water, between freezing and boiling temperatures and at atmospheric pressure.
After mixing the water, the total of 150 grams will reach an equilibrium temperature T.
The water originally at 60 degrees temperature will "loose" an amount of heat equal to: Delta Q = c*100 [grams]*(60-T) [degrees].
This same amount of heat will be absorbed by the water originally at 40 degrees to raise its temperature to T. Delta Q = c*50*(T-40).
Now it is easy to solve this equality:
c*100*(60-T) = c*50*(T-40), giving T=53⅓ degrees Celsius.
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