A piece of copper metal is initially at 100.0°C. It is dropped into a coffee cup calorimeter containing 50.0 g of water at a temperature of 20.0°C. After stirring, the final temperature of both copper and water is 25.0°C. Assuming no heat losses, and that the specific heat (capacity) of water is 4.184 J/(g•K), what is the heat capacity of the copper & the calorimeter in J/K?
3.19
5665
To find the heat capacity of the copper and the calorimeter, we need to calculate the heat gained by the water and the copper.
The heat gained by the water can be calculated using the formula:
Q_water = m_water * c_water * ΔT_water
Where:
Q_water = heat gained by water
m_water = mass of water (50.0 g)
c_water = specific heat capacity of water (4.184 J/(g·K))
ΔT_water = change in temperature of water (final temperature - initial temperature)
Substituting the given values:
Q_water = 50.0 g * 4.184 J/(g·K) * (25.0°C - 20.0°C)
Q_water = 5230 J
Since energy is conserved, the heat gained by the water is equal to the heat lost by the copper and the calorimeter, so:
Q_copper_calorimeter = Q_water
Now, let's calculate the heat capacity of the copper and the calorimeter.
The heat capacity is defined as the amount of heat required to raise the temperature of an object by 1 degree Kelvin or Celsius. Therefore, the heat capacity of the copper and the calorimeter can be calculated using the formula:
Heat capacity = Q / ΔT
Where:
Heat capacity = heat capacity of the copper/calorimeter
Q = heat gained/lost by the copper/calorimeter
ΔT = change in temperature of the copper/calorimeter (final temperature - initial temperature)
For the copper and calorimeter together, the heat capacity (Copper + Calorimeter) is:
Heat capacity (Copper + Calorimeter) = Q_copper_calorimeter / ΔT_copper_calorimeter
From the given data:
Q_copper_calorimeter = 5230 J
ΔT_copper_calorimeter = (25.0°C - 100.0°C) = -75.0°C
Plugging in these values, we can calculate the heat capacity:
Heat capacity (Copper + Calorimeter) = 5230 J / -75.0°C
Heat capacity (Copper + Calorimeter) = -69.73 J/°C
Therefore, the heat capacity of the copper and the calorimeter is -69.73 J/°C.
To find the heat capacity of the copper and the calorimeter, we need to use the principle of conservation of energy. The heat lost by the copper piece will be equal to the heat gained by the water and the calorimeter.
The equation we can use is:
Heat lost by copper = Heat gained by water + Heat gained by calorimeter
Let's calculate each term one by one:
1. Heat lost by copper:
The formula to calculate heat is:
Q = mcΔT
where Q is the heat, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
The mass of the copper piece is not given, but we can assume it to be 1 gram for simplicity.
So, for the copper:
Q(copper) = mcΔT
Q(copper) = 1g * c(copper) * (T(final) - T(initial))
2. Heat gained by water:
For the water:
Q(water) = mw * c(water) * (T(final) - T(initial))
Q(water) = 50g * 4.184 J/(g·K) * (25.0°C - 20.0°C)
3. Heat gained by the calorimeter:
Since the calorimeter is the container in which the water is placed, it gains heat too. The heat capacity of the calorimeter is assumed to be negligible compared to the water.
Therefore, we can assume that Q(calorimeter) is equal to Q(water).
Now, we can substitute the values in the equation:
Q(copper) = Q(water) + Q(calorimeter)
1g * c(copper) * (T(final) - T(initial)) = 50g * 4.184 J/(g·K) * (25.0°C - 20.0°C)
Simplifying the equation will give us the value of c(copper), the specific heat capacity of copper and the calorimeter in J/(g·K).