A student conducted a calorimetry experiment where 76.0 g of silver metal at 75.0°C was dropped into a Styrofoam cup
calorimeter containing 125 g of water at 15.0°C. Also in the cup was a 35.0 g glass stirring rod, also at 15.0°C. Assuming no heat is transferred between the system & the Styrofoam cup, what was the final temperature of the system?
You need to know the specific heat of the Ag metal.
You need to know the specific heat of the glass rod.
You need to know the specific heat of water.
[mass Ag x specific heat Ag x (Tfinal-Tinital)] + [mass of glass rod x specific heat glass rod x (Tfinal-Tintial)] + [mass water x specific heat water x (Tfinal-Tinitial)] = 0
Substitute the numbers and solve for Tfinal. Post your work if you get stuck.
To find the final temperature of the system, we can use the principle of heat transfer. The heat lost by the silver metal when it cools down will be equal to the heat gained by the water, glass stirring rod, and the Styrofoam cup.
Let's calculate the heat lost by the silver metal first:
Qsilver = msilver * csilver * ΔTsilver
Where,
msilver = mass of silver metal = 76.0 g
csilver = specific heat capacity of silver = 0.235 J/g°C (given in the question)
ΔTsilver = change in temperature of the silver metal = final temperature - initial temperature of silver metal = final temperature - 75.0°C
The heat gained by the water, glass stirring rod, and the Styrofoam cup will be:
Qwater + Qglass + Qcup = mwater * cwater * ΔTwater + mglass * cglass * ΔTglass + mcup * ccup * ΔTcup
Where,
mwater = mass of water = 125 g
cwater = specific heat capacity of water = 4.18 J/g°C (given in the question)
ΔTwater = change in temperature of water = final temperature - initial temperature of water = final temperature - 15.0°C
mglass = mass of glass stirring rod = 35.0 g
cglass = specific heat capacity of glass = 0.840 J/g°C (assuming it is made of borosilicate glass)
ΔTglass = change in temperature of glass stirring rod = final temperature - initial temperature of glass stirring rod = final temperature - 15.0°C
mcup = mass of Styrofoam cup = mass of water + mass of glass stirring rod = 125 g + 35.0 g = 160 g
ccup = specific heat capacity of Styrofoam cup (approximated as water, assuming no heat transfer) = 4.18 J/g°C (given in the question)
ΔTcup = change in temperature of Styrofoam cup = final temperature - initial temperature of Styrofoam cup = final temperature - 15.0°C
Since there is no heat transferred between the system and the Styrofoam cup, the heat lost by the silver metal must be equal to the heat gained by the water, glass stirring rod, and the Styrofoam cup:
msilver * csilver * ΔTsilver = mwater * cwater * ΔTwater + mglass * cglass * ΔTglass + mcup * ccup * ΔTcup
Now, we can substitute the given values into the equation and solve for the final temperature.
To find the final temperature of the system, we need to consider the principle of conservation of energy. The heat lost by the silver metal and the glass stirring rod should be equal to the heat gained by the water.
First, we need to calculate the heat lost by the silver metal and the glass stirring rod using the formula:
Q = mcΔT
where:
Q is the heat lost or gained,
m is the mass of the substance,
c is the specific heat capacity of the substance, and
ΔT is the change in temperature.
For silver metal:
m = 76.0 g (mass of silver metal)
c = 0.24 J/g°C (specific heat capacity of silver)
ΔT = 75.0°C - final temperature (since the initial temperature of the silver metal is not known)
The heat lost by the silver metal is given by:
Q1 = (76.0 g) * (0.24 J/g°C) * (75.0°C - final temperature)
For the glass stirring rod:
m = 35.0 g (mass of glass stirring rod)
c = 0.84 J/g°C (specific heat capacity of glass)
ΔT = 15.0°C - final temperature
The heat lost by the glass stirring rod is given by:
Q2 = (35.0 g) * (0.84 J/g°C) * (15.0°C - final temperature)
According to the principle of conservation of energy, the heat lost by the silver metal and the glass stirring rod is equal to the heat gained by the water.
Q1 + Q2 = (125 g) * (4.18 J/g°C) * (final temperature - 15.0°C)
We can equate the two equations and solve for the final temperature:
(76.0 g) * (0.24 J/g°C) * (75.0°C - final temperature) + (35.0 g) * (0.84 J/g°C) * (15.0°C - final temperature) = (125 g) * (4.18 J/g°C) * (final temperature - 15.0°C)
Simplifying and solving this equation will give us the final temperature of the system.