2. A 75.0 g sample of a metal is placed in boiling water until its temperature is 100.00C. A calorimeter contains 100.00 g of water at a temperature of 24.40C. The metal sample is removed from the boiling water and immediately place in water in the calorimeter. The final temperature of the metal and water in the calorimeter is 34.90C. Assuming that the calorimeter provides perfect insulation, what is the specific heat of the metal? The specific heat of water is 4.184 J/(g•0C).
To find the specific heat of the metal, we can use the principle of conservation of energy. The heat gained by the water in the calorimeter is equal to the heat lost by the metal.
The heat gained by the water can be calculated using the formula:
Q_water = m_water * C_water * ΔT_water
Where:
m_water is the mass of water in grams (100.00 g),
C_water is the specific heat of water (4.184 J/(g•°C)),
ΔT_water is the change in temperature of water (final temperature - initial temperature)
Q_water = 100.00 g * 4.184 J/(g•°C) * (34.90°C - 24.40°C)
The heat lost by the metal is given by:
Q_metal = m_metal * C_metal * ΔT_metal
Where:
m_metal is the mass of the metal in grams (75.0 g),
C_metal is the specific heat of the metal (unknown),
ΔT_metal is the change in temperature of the metal (initial temperature - final temperature)
Since we are assuming the calorimeter provides perfect insulation, the heat lost by the metal will be gained by the water in the calorimeter. Therefore, we can set Q_water equal to Q_metal:
m_water * C_water * ΔT_water = m_metal * C_metal * ΔT_metal
Plugging in the values we know:
100.00 g * 4.184 J/(g•°C) * (34.90°C - 24.40°C) = 75.0 g * C_metal * (100.00°C - 34.90°C)
Simplifying the equation:
4184 J * 10.50°C = 75.0 g * C_metal * 65.10°C
Cancelling out the grams:
4184 J * 10.50°C = C_metal * 75.0 * 65.10°C
Dividing both sides by (75.0 * 65.10 °C):
C_metal = (4184 J * 10.50°C) / (75.0 * 65.10 °C)
C_metal = 5691.6 J / (75.0 * 65.10 °C)
C_metal ≈ 1.115 J/(g•°C)
Therefore, the specific heat of the metal is approximately 1.115 J/(g•°C).
To find the specific heat of the metal, we can use the principle of conservation of energy. We can assume that there is no heat loss to the surroundings since the calorimeter provides perfect insulation.
The energy gained by the metal when it cools down is equal to the energy lost by the water when it heats up. We can calculate these energies using the formula:
Q = mcΔT
where:
Q is the energy in joules (J)
m is the mass in grams (g)
c is the specific heat in J/(g•°C)
ΔT is the change in temperature in °C
For the metal, we have:
Q_metal = m_metal * c_metal * ΔT_metal = -Q_water
For the water, we have:
Q_water = m_water * c_water * ΔT_water
Since the calorimeter provides perfect insulation, the result should be negative for the water, indicating a loss of energy.
Substituting the known values from the problem:
m_metal = 75.0 g
c_water = 4.184 J/(g•°C)
ΔT_metal = (34.90°C - 24.40°C) = 10.50°C
ΔT_water = (34.90°C - 24.40°C) = 10.50°C
Now we can substitute these values into the equations we derived earlier:
Q_metal = 75.0 g * c_metal * 10.50°C
Q_water = 100.00 g * 4.184 J/(g•°C) * 10.50°C
Since the two quantities are equal when one is positive and the other is negative, we can set them equal to each other and solve for c_metal:
Q_metal = Q_water
75.0 g * c_metal * 10.50°C = -100.00 g * 4.184 J/(g•°C) * 10.50°C
Simplifying the equation:
c_metal = -100.00 g * 4.184 J/(g•°C) * 10.50°C / (75.0 g * 10.50°C)
Now, calculate c_metal:
c_metal = -100.00 g * 4.184 J/(g•°C) / 75.0 g
c_metal = -418.4 J/°C / 75.0
c_metal = -5.58 J/g•°C
Therefore, the specific heat of the metal is approximately -5.58 J/(g•°C). Note that the negative sign indicates that the metal has a different sign convention for heat compared to water.
Use the same formula for the next question up, substitute, solve for specific heat metal.
Post your work if you get stuck.