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?
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