A 60-g ice cube at 0°C is placed in 800 g of water at 30°C. What is the final temperature of the mixture?
A 60-g ice cube at 0°C is placed in 800 g of water at 30°C. What is the final temperature of the mixture?
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q = mcΔT
> q = heat
> m = mass
> c = specific heat
> ΔT = final temperature - initial temperature
the heat gained by the ice = the heat lost by the water
q(ice) = -q(water)
mcΔT(ice) = -mcΔT(water)
Plug and chug. Solve for final temperature. Specific heat of water is 4.184. Tf is final temperature.
(60)(4.184)(Tf - 0) = -(800)(4.184)(Tf - 30)
(60)(Tf - 0) = (-800)(Tf - 30)
60Tf = -800Tf + 24000
860Tf = 24000
Tf = 27.9C
Final temperature is 27.9C