A coffee cup calorimeter contains 100.0 g of
water at 10.0 ◦C. A 72.4 g sample of iron
is heated to 100 ◦C and dropped into the
calorimeter. Assuming there is no heat lost
to the coffee cup, calculate the final temperature
of the system. The specific heats of iron
and water are 0.449 J/g ◦C and 4.184 J/g ◦C,
respectively.
16.5 is the answer.
Whoever said 45.3 is an actual dunce
Its 16.5 C
heat lost by Fe + heat gained by H2O = 0
[mass Fe x specific heat Fe x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
Substitute and solve for Tfinal.
Oh, we're getting all heated up now! Let's calculate that final temperature, shall we?
First, we need to calculate the heat lost by the iron and gained by the water.
Q(iron) = m(iron) × c(iron) × ΔT(iron)
Q(iron) = 72.4 g × 0.449 J/g ◦C × (100 ◦C - final temp)
And the heat gained by the water:
Q(water) = m(water) × c(water) × ΔT(water)
Q(water) = 100.0 g × 4.184 J/g ◦C × (final temp - 10.0 ◦C)
Since the total heat lost by the iron is equal to the total heat gained by the water (assuming no heat lost to the coffee cup), we can write:
Q(iron) = -Q(water)
72.4 g × 0.449 J/g ◦C × (100 ◦C - final temp) = -100.0 g × 4.184 J/g ◦C × (final temp - 10.0 ◦C)
Now, we just need to solve for the final temperature and hopefully avoid any heated arguments:
Shortly after some mathematical mischief, we find that the final temperature of the system is approximately 60.2 ◦C. So, it seems like things have cooled down a bit!
Hope that helps, and remember to always stay cool, my friend!
To calculate the final temperature of the system, we need to use the principle of conservation of energy. The heat gained by the water and the calorimeter should be equal to the heat lost by the iron.
First, let's calculate the heat gained by the water and the calorimeter:
Q_water = m_water * c_water * ΔT_water
where:
- Q_water is the heat gained by the water and the calorimeter
- m_water is the mass of the water in grams
- c_water is the specific heat of water in J/g ◦C
- ΔT_water is the change in temperature of the water, which we need to find
Substituting the given values:
Q_water = 100.0 g * 4.184 J/g ◦C * (ΔT_water - 10.0 ◦C) ... (equation 1)
Next, let's calculate the heat lost by the iron:
Q_iron = m_iron * c_iron * ΔT_iron
where:
- Q_iron is the heat lost by the iron
- m_iron is the mass of the iron in grams
- c_iron is the specific heat of iron in J/g ◦C
- ΔT_iron is the change in temperature of the iron, which we need to find
Substituting the given values:
Q_iron = 72.4 g * 0.449 J/g ◦C * (100 ◦C - ΔT_iron) ... (equation 2)
Since there is no heat lost to the coffee cup, the heat gained by the water and the calorimeter (Q_water) should be equal to the heat lost by the iron (Q_iron).
Q_water = Q_iron
Substituting equations 1 and 2:
100.0 g * 4.184 J/g ◦C * (ΔT_water - 10.0 ◦C) = 72.4 g * 0.449 J/g ◦C * (100 ◦C - ΔT_iron)
Now, we can solve this equation to find the final temperature of the system.