A 5.00 g sample of Al pellets (specific heat = .89 J/g degrees c) and a 10.00 g sample of Fe pellets (specific heat = .45 J/g degrees c) are heated to 100.0 degrees C. The mixture of hot Fe and Al is then dropped into 97.3 g of water at 22.0 degrees C. Calculate the final temp of the metal and water mixture, assuming no heat is lost to surroundings.
how do i do this problem? i'm completely lost.
The sum of the heats gained is zero (so will be absorbing heat, so will be negative heat gain).
HeatgainedAl + heatgainedFe + heatgainedwater=0
HeatgainedAl= massAl*Ca*(tf-100)
heat gainedFe=massFe*Cfe*(tf-100)
heatgainedwater=masswater*ch2o*(tf-22)
solve for tf.
how do i solve for tf if there's nothing on the other side of the equal sign?
To solve this problem, you need to use the principle of conservation of energy, which states that the heat gained by the water and the heat lost by the metals will be equal when they reach thermal equilibrium.
First, let's calculate the heat gained or lost by the aluminum (Al) and iron (Fe) pellets when they cool down from 100.0°C to the final temperature Tf.
QAl = mAl × cAl × ΔT
= 5.00 g × 0.89 J/g°C × (Tf - 100.0°C)
QFe = mFe × cFe × ΔT
= 10.00 g × 0.45 J/g°C × (Tf - 100.0°C)
Now, considering that the heat lost by Al and Fe is equal to the heat gained by the water, we can set up an equation:
QAl + QFe = Qwater
Substituting the above equations, we have:
5.00 g × 0.89 J/g°C × (Tf - 100.0°C) + 10.00 g × 0.45 J/g°C × (Tf - 100.0°C) = 97.3 g × 4.18 J/g°C × (Tf - 22.0°C)
Simplifying the equation, we get:
4.45 Tf - 445.00 + 9.00 Tf - 900.00 = 406.514 Tf - 9001.36
Combining like terms:
13.45 Tf - 1345.00 = 406.514 Tf - 9001.36
Re-arranging the terms:
393.064 Tf = 7656.36
Solving for Tf:
Tf = 19.49°C
So, the final temperature of the metal and water mixture is approximately 19.49°C.