What is the final temperature reached by the addition of 11.4 grams of steam at 100C to 681 grams water at 25C in an aluminum calorimeter having a mass of 781 grams?
C of water=1 cal/gc
C of aluminum= 0.22 cal/gc
L steam= 540cal/g
My Attempt
Qloss=Qgain
Qst+Qhw=Qcup+Qcw
ml+mcT=mcT+mcT
after plugging the numbers I'm not sure. All three final temps are the same at equilibrium and the cup and cold water start at 20. The hot water on the steam side starts at 100c. How do I solve? Thank you very much!!
I don't see how you handled the heat of vaporization.
Sum of heats gained=0
11.4(-Hv)+681cwater*(Tf-25)+781cAl*(tf-100)=0
-11.4*540+681*1*(Tf-25)+781*.22*(Tf-100)=0
do the algebra, solve for Tf.
Thanks for answering so quickly. I know there are different ways to doing things and I'm by no means an expert but don't we also have to account for the water that was previously steam, in this case 11.4 grams of water?
To find the final temperature, you can use the principle of energy conservation. The equation you wrote, Qloss = Qgain, is correct.
Let's break down the equation and solve step by step:
Qst: The heat gained by the steam can be calculated using the formula Q = mL, where m is the mass of the steam and L is the specific heat of steam.
Qhw: The heat gained by the hot water can be calculated using the formula Q = mCΔT, where m is the mass of the hot water, C is the specific heat of water, and ΔT is the change in temperature.
Qcup: The heat lost by the calorimeter can be calculated using the formula Q = mcΔT, where m is the mass of the calorimeter, c is the specific heat of aluminum, and ΔT is the change in temperature.
Qcw: The heat lost by the cold water can be calculated using the formula Q = mCΔT, where m is the mass of the cold water, C is the specific heat of water, and ΔT is the change in temperature.
Now, you can plug in the given values into the equation and solve for the final temperature:
Qst + Qhw = Qcup + Qcw
(mL) + (mCΔT) = (mcΔT) + (mCΔT)
Substituting the given values:
(11.4g)(540 cal/g) + (11.4g)(1 cal/g°C)(T-100°C) = (781g)(0.22 cal/g°C)(T-25°C) + (681g)(1 cal/g°C)(T-25°C)
Now, you can solve for T. Start by simplifying the equation and combining like terms:
6156 + 11.4(T-100) = 171.82(T-25) + 681(T-25)
6156 + 11.4T -1140 = 171.82T - 4295.5 + 681T - 17025.5
Combine like terms again:
11.4T - 1140 = 852.82T - 21321
Now, isolate the T term by bringing all the T terms to one side and constant terms to the other side:
11.4T - 852.82T = -21321 + 1140
-841.42T = -20181
Finally, solve for T:
T = -20181 / -841.42
T ≈ 23.98°C
Therefore, the final temperature reached by the system is approximately 23.98°C.