Assume you are performing the calibration step of Experiment 8 and you begin with 40 g of water at 20 oC and 40 g of water at 80 oC. After adding the two portions of water into your calorimeter setup and following the procedure outlined in the experiment, you determine the temperature of the mixed portions of water to be 45 oC. What is the heat capacity of the calorimeter. Assume room temperature is 25 oC.
To find the heat capacity of the calorimeter, we can use the principle of energy conservation. The heat gained by the colder water is equal to the heat lost by the hotter water. We can calculate the heat gained or lost using the formula:
Q = mcΔT
Where Q represents heat, m represents mass, c represents specific heat capacity, and ΔT represents the change in temperature.
In this case, we have two portions of water, each with a mass of 40 g. The temperature of the cold water is 20 oC and the temperature of the hot water is 80 oC. The final temperature, after mixing, is 45 oC. We can calculate the heat gained or lost by each portion of water individually and set them equal to each other to solve for the calorimeter's heat capacity.
Heat gained by the cold water:
Q1 = mcΔT1
= (40 g) * (1 cal/g oC) * (45 oC - 20 oC)
= 40 cal * 25 oC
= 1000 cal
Heat lost by the hot water:
Q2 = mcΔT2
= (40 g) * (1 cal/g oC) * (45 oC - 80 oC)
= 40 cal * (-35 oC)
= -1400 cal
Since the heat gained by the cold water is equal to the heat lost by the hot water, we can equate the two equations:
Q1 = -Q2
1000 cal = -(-1400 cal)
1000 cal = 1400 cal
Now, we know that the heat capacity of the calorimeter is equal to the heat gained or lost divided by the change in temperature:
Ccal = Q/ΔT
= (1000 cal)/(45 oC - 25 oC)
= 1000 cal/20 oC
= 50 cal/oC
Therefore, the heat capacity of the calorimeter is 50 cal/oC.
To find the heat capacity of the calorimeter, we can use the principle of heat transfer, which states that the heat lost by the hot water is equal to the heat gained by the cold water and the calorimeter. We can write this as:
Heat lost by hot water = Heat gained by cold water + Heat gained by the calorimeter
Since we know the mass and initial and final temperatures of the water portions, we can calculate the heat lost by the hot water using the formula:
Q = mcΔT
Where:
Q is the heat lost
m is the mass of the hot water
c is the specific heat capacity of water
ΔT is the change in temperature of the hot water
Given:
Mass of hot water (m_hot) = 40 g
Initial temperature of hot water (T_hot_initial) = 80 °C
Final temperature of the mixed water (T_mixed) = 45 °C
Specific heat capacity of water (c) = 4.18 J/g°C
First, calculate the heat lost by the hot water:
Q_hot = m_hot * c * (T_hot_initial - T_mixed)
Q_hot = 40 g * 4.18 J/g°C * (80 °C - 45 °C)
Q_hot = 40 g * 4.18 J/g°C * 35 °C
Q_hot = 5866 J
Next, we need to calculate the heat gained by the cold water using the same formula:
Q_cold = m_cold * c * (T_mixed - T_cold_initial)
Since the mass of the cold water is the same as the mass of the hot water, and the specific heat capacity of water is the same, we can directly substitute the values:
Q_cold = Q_hot
Finally, we can calculate the heat gained by the calorimeter:
Q_calorimeter = Q_cold - Q_hot
Since Q_cold = Q_hot, the heat gained by the calorimeter is zero.
Therefore, the heat capacity of the calorimeter is zero, assuming no heat is gained or lost to the surroundings.