A copper calorimeter of mass ()g contains 380g of a liquid at 12degrees Celsius. A 20W heater is used for 3 minutes to raise the temperature of the liquid and calorimeter to 17 degrees Celsius. The calorimeter has a specific heat capacity of 400 J/K kg
a.Calculate the heat energy supplied to the calorimeter and the liquid inside it.
b.Calculate the heat supplied to the calorimeter.
c.Assuming no heat losses find the specific heat capacity of the liquid....
I just couldn't do part b and c. Is there enough information supplied in the question? Thanks
To solve part b, we need to calculate the heat supplied to the calorimeter. The formula to calculate heat is:
Heat (Q) = mass (m) * specific heat capacity (c) * change in temperature (ΔT)
In this case, the mass of the calorimeter is not given directly, but we can assume it to be the same as the mass of the liquid since they are both inside the calorimeter. The specific heat capacity of the calorimeter (c) is also given as 400 J/K kg.
We know that the initial temperature (T1) of the liquid and calorimeter is 12 degrees Celsius, and the final temperature (T2) is 17 degrees Celsius. So, the change in temperature (ΔT) is 17 - 12 = 5 degrees Celsius.
Now, let's calculate the heat supplied to the calorimeter using the formula:
Q = m * c * ΔT
Substituting the values:
Q = 380g * 400 J/K kg * 5 degrees Celsius
To calculate the heat supplied to the calorimeter, we need to convert the mass from grams to kilograms:
380g = 380/1000 kg = 0.38 kg
Now, solving for Q:
Q = 0.38 kg * 400 J/K kg * 5 degrees Celsius
Q = 760 J
Therefore, the heat supplied to the calorimeter is 760 J.
Now, let's move on to part c. To calculate the specific heat capacity of the liquid, we need to assume no heat losses, which means all the heat supplied to the calorimeter is transferred to the liquid.
We already know that the heat supplied to the calorimeter is 760 J, and the mass of the liquid inside the calorimeter is 380 g.
Using the same formula as before:
Q = m * c * ΔT
760 J = 380g * c * 5 degrees Celsius
To calculate the specific heat capacity of the liquid (c), we need to rearrange the formula:
c = Q / (m * ΔT)
Now, substituting the values:
c = 760 J / (380g * 5 degrees Celsius)
Again, converting the mass from grams to kilograms:
380g = 380/1000 kg = 0.38 kg
c = 760 J / (0.38 kg * 5 degrees Celsius)
Simplifying further:
c = 760 J / (1.9 kg * 5 degrees Celsius)
c = 760 J / 9.5 K kg
c = 80 J/K kg
Therefore, assuming no heat losses, the specific heat capacity of the liquid is 80 J/K kg.
So, to summarize:
a. The heat energy supplied to the calorimeter and the liquid inside it is 760 J.
b. The heat supplied to the calorimeter is 760 J.
c. Assuming no heat losses, the specific heat capacity of the liquid is 80 J/K kg.
Note that these calculations assume that the specific heat capacity of the liquid is constant over this temperature range and that the calorimeter is well-insulated to minimize heat losses.