Asked by Dale
how many grams of water at 20 degrees celcius are necessary to change 800.0g of water at 90.0 degrees celsius to 50.0 degrees celcius?
Answers
Answered by
Elena
Q= mc Δt
mc(50-20) = Mc(90-50)
m=M•40/30 = 800•40/30 = 1067 kg
mc(50-20) = Mc(90-50)
m=M•40/30 = 800•40/30 = 1067 kg
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