A 30.5 g sample of an alloy at 94.0°C is placed into 48.7 g water at 20.3°C in an insulated coffee cup. The heat capacity of the coffee cup (without the water) is 9.2 J/K. If the final temperature of the system is 31.1°C, what is the specific heat capacity of the alloy? (c of water is 4.184 J/g×K)
[mass alloy x specific heat alloy x (Tfinal-Tinitial)] + [(calorimeter constant x (Tfinal-Tinitial)] = 0
(30.5)(X)(10.8)+(?)(10.8)=0
What number goes in the calorimeter constant ?
9.2 J/K = heat capacity of the coffee cup.
It's still not giving me the right answer. I'm pulling in the numbers, and doing the equation right. But, it keeps telling me, I am wrong.
I didn't include the water and should have done so.
(mass water x specific heat water x (Tfinal-Tinitial)) + (mass alloy x specific heat alloy x (Tfinal-Tinitial)) + Ccal(Tfinal-Tinitial) = 0
so, what is the t final- t initial for ccal?
To find the specific heat capacity of the alloy, we can use the principle of heat transfer. The heat gained by the alloy and the water should be equal to the heat lost by the coffee cup.
The heat gained by the alloy can be calculated using the formula:
Q_alloy = m_alloy * c_alloy * ΔT_alloy
Where:
Q_alloy is the heat gained by the alloy
m_alloy is the mass of the alloy (30.5 g)
c_alloy is the specific heat capacity of the alloy (unknown)
ΔT_alloy is the change in temperature of the alloy (final temperature - initial temperature)
The heat gained by the water can be calculated using the formula:
Q_water = m_water * c_water * ΔT_water
Where:
Q_water is the heat gained by the water
m_water is the mass of the water (48.7 g)
c_water is the specific heat capacity of water (4.184 J/g×K)
ΔT_water is the change in temperature of the water (final temperature - initial temperature)
Since the coffee cup is insulated, it does not gain or lose any heat, so the heat lost by the cup is zero:
Q_cup = 0
The total heat gained is equal to the total heat lost:
Q_alloy + Q_water = Q_cup
Substituting the formulas and values:
m_alloy * c_alloy * ΔT_alloy + m_water * c_water * ΔT_water = 0
We are given the values for m_alloy (30.5 g), m_water (48.7 g), ΔT_water (31.1°C - 20.3°C), c_water (4.184 J/g×K), and ΔT_alloy as the final temperature (31.1°C) minus the initial temperature (94.0°C).
Using this equation, we can solve for c_alloy:
c_alloy = - (m_water * c_water * ΔT_water) / (m_alloy * ΔT_alloy)
Plugging in the values and calculating:
c_alloy = - (48.7 g * 4.184 J/g×K * (31.1°C - 20.3°C)) / (30.5 g * (31.1°C - 94.0°C))
c_alloy = - (48.7 g * 4.184 J/g×K * 10.8°C) / (30.5 g * (-62.9°C))
c_alloy = - (217.201 J * °C) / (-1941.45 J)
c_alloy = 0.1118 J/°C
Therefore, the specific heat capacity of the alloy is approximately 0.1118 J/°C.