A 50g chunk of metal (c=0.0506 j/g°C) is heated to 60°C. it is then dropped into a calorimeter(c=1.40 j/g°C,m=5.0 g)that contains 100.0g of an unknown liquid. the temperature of the liquid raises from 15°C to 20°C.What is the c of the calorimeter?
Isn't c given in the problem for the calorimeter as 1.40 J/g*C?
It's supposed to be "c" of liquid ,I'm sorry.
heat capacity of the calorimeter is 1.40 J/g*C x 5 g = 7.00 J/C
heat lost by metal + heat gained by calorimeter + heat gained by water = 0
[mass metal x specific heat metal x (Tfinal-Tinitial)] + [7.00 x (Tfinal-Tinitial)] + [mass liquid x specific heat liquid x (Tfinal-Tinitial)] = 0
note:Tf and Ti are the same for the calorimeter and liquid since the liquid is IN the calorimeter.
Substitute and solve for specific heat liquid.
To find the specific heat capacity (c) of the calorimeter, we can use the principle of energy conservation. The heat lost by the metal when it cools down is equal to the heat gained by the unknown liquid and the calorimeter.
First, let's calculate the heat gained by the unknown liquid. We can use the formula:
Q = mcΔT
where Q is the heat gained, m is the mass of the unknown liquid, c is its specific heat capacity, and ΔT is the change in temperature.
The mass of the unknown liquid is given as 100.0g, the change in temperature is from 15°C to 20°C, and the specific heat capacity is unknown. Let's denote the specific heat capacity of the unknown liquid as c_liquid.
Q_liquid = m * c_liquid * ΔT_liquid
= 100.0g * c_liquid * (20°C - 15°C)
= 500.0g * c_liquid
Now, let's calculate the heat gained by the calorimeter. Similar to the previous calculation, we can use the same formula:
Q_calorimeter = mcΔT
This time, m represents the mass of the calorimeter (given as 5.0g), c represents its specific heat capacity (unknown, denoted as c_calorimeter), and ΔT represents the change in temperature (which we can calculate as the difference between the initial and final temperatures of the liquid).
Q_calorimeter = 5.0g * c_calorimeter * ΔT_calorimeter
Now, since energy is conserved, the heat lost by the metal is equal to the heat gained by the unknown liquid and the calorimeter:
Q_metal = Q_liquid + Q_calorimeter
The heat lost by the metal can be calculated using the specific heat capacity of the metal (given as c = 0.0506 J/g°C) and the change in temperature (from 60°C to the final temperature of the liquid).
Q_metal = 50g * 0.0506 J/g°C * (60°C - Tf)
where Tf is the final temperature of the liquid.
Now, equating the two equations above, we get:
50g * 0.0506 J/g°C * (60°C - Tf) = 500.0g * c_liquid + 5.0g * c_calorimeter * ΔT_calorimeter
We also know that ΔT_calorimeter = Tf - 20°C (since the calorimeter's temperature change is from 20°C to Tf).
Substituting the value of ΔT_calorimeter in the equation, we can simplify it to find c_calorimeter:
50g * 0.0506 J/g°C * (60°C - Tf) = 500.0g * c_liquid + 5.0g * c_calorimeter * (Tf - 20°C)
Solving this equation will give us the value of c_calorimeter, which represents the specific heat capacity of the calorimeter.