A student calibrates his/her calorimeter. S/he performs part 1 as instructed. The data are: 99.2 g water at 36.0°C are added to 97.3 g water in the calorimeter at 16.0°C. The final temperature is 26.0°C. Calculate the heat capacity of the calorimeter in J/°C.
To calculate the heat capacity of the calorimeter, we need to use the equation:
q = m * c * ΔT
Where:
- q is the heat gained or lost by the system (calorimeter + water)
- m is the mass of the substance (water)
- c is the specific heat capacity of the substance (water)
- ΔT is the change in temperature (final temperature - initial temperature)
First, let's calculate the heat gained or lost by the system.
To find the mass of the water that is mixed, we add the masses of the two quantities given: 99.2 g + 97.3 g = 196.5 g
Next, we calculate the change in temperature: ΔT = 26.0°C - ((36.0°C + 16.0°C) / 2) = 26.0°C - 26.0°C = 0°C
Since the temperature change is 0°C, this means that no heat was lost or gained by the water.
Now, let's calculate the heat capacity of the calorimeter.
Since no heat was transferred to or from the water, the heat gained by the water must be equal to the heat lost by the calorimeter.
Therefore,
q_calorimeter = -q_water
Plugging in the values:
m_calorimeter * c_calorimeter * ΔT_calorimeter = -(m_water * c_water * ΔT_water)
We know the mass of the water (196.5 g), specific heat capacity of water (4.18 J/g°C), and the change in temperature of the water (0°C).
We only need to solve for the heat capacity of the calorimeter (c_calorimeter).
m_calorimeter * c_calorimeter * ΔT_calorimeter = -(m_water * c_water * ΔT_water)
c_calorimeter = -(m_water * c_water * ΔT_water) / (m_calorimeter * ΔT_calorimeter)
c_calorimeter = -(196.5 g * 4.18 J/g°C * 0°C) / (97.3 g * (26.0 - (36.0 + 16.0) / 2)°C)
c_calorimeter = 0 J/°C
Therefore, the heat capacity of the calorimeter is 0 J/°C.
To calculate the heat capacity of the calorimeter, we need to use the equation:
q = m * c * ΔT
Where:
q = heat transferred
m = mass of the substance (in this case, water)
c = specific heat capacity of the substance (in this case, water)
ΔT = change in temperature
In the given situation, we have two parts:
Part 1: 99.2 g water at 36.0°C added to 97.3 g water at 16.0°C
First, let's calculate the heat transferred in Part 1:
q1 = m1 * c1 * ΔT1
q1 = (99.2 g + 97.3 g) * c_water * (26.0°C - ((36.0°C + 16.0°C) / 2))
The average temperature [(36.0°C + 16.0°C) / 2] is used for ΔT1.
To calculate the specific heat capacity of water (c_water), we can use the value 4.18 J/g°C.
Now, let's substitute the values into the equation:
q1 = (196.5 g) * (4.18 J/g°C) * (26.0°C - 26.0°C)
Since the water in Part 1 is not raising or lowering in temperature, the value of q1 will be zero, as there is no heat transfer.
Therefore, q1 = 0 J.
Now, let's move to Part 2: the calorimeter.
In Part 2, the calorimeter is heated from 16.0°C to 26.0°C using the heat transferred from Part 1. This transfer of heat is equivalent to the heat capacity of the calorimeter.
q2 = m2 * c_calorimeter * ΔT2
The mass of the calorimeter is not given, but we can assume it to be a negligible value compared to the mass of the water.
Now, let's substitute the values into the equation:
q2 = m2 * c_calorimeter * (26.0°C - 16.0°C)
Because q1 and q2 are equal (assuming no heat is lost to the surroundings), we can equate the two equations:
q1 = q2
0 J = m2 * c_calorimeter * (26.0°C - 16.0°C)
Simplifying the equation:
0 J = m2 * c_calorimeter * 10.0°C
Since m2 is not given, we can assume it to be 1 g for the purposes of this calculation.
0 J = 1 g * c_calorimeter * 10.0°C
c_calorimeter = 0 J / (1 g * 10.0°C)
Since the value of q1 is 0 J, the heat capacity of the calorimeter would be 0 J/°C.
Hence, the heat capacity of the calorimeter is 0 J/°C.