A piece of copper ball of mass 20g at200 Is placed in a copper calorimeter of mass 60g containing 50g of water at 30 ignoring the heat loss calculate the final temperaure of the mixture
T is between 200 deg and 30 deg
heat out of copper = Cc * 20 ( 200-T)
heat into water = Cw * 50 (T-30)
they are equal
Good question
To calculate the final temperature of the mixture, we will use the principle of conservation of energy.
The heat gained by the water and the calorimeter equals the heat lost by the copper ball.
Heat gained by water and calorimeter = Heat lost by copper ball
The heat gained by the water and calorimeter can be calculated using the formula:
Q = m * c * ΔT
Where:
Q = heat gained by water and calorimeter
m = mass (water + calorimeter)
c = specific heat capacity of water
ΔT = change in temperature
For water, the specific heat capacity (c) is approximately 4.18 J/g°C.
The heat lost by the copper ball can be calculated using the formula:
Q = m * c * ΔT
Where:
Q = heat lost by copper ball
m = mass of the copper ball
c = specific heat capacity of copper
For copper, the specific heat capacity (c) is approximately 0.39 J/g°C.
Since the heat gained and lost are equal, we can set up an equation:
(m_water + m_calorimeter) * c_water * ΔT_water = m_copper * c_copper * ΔT_copper
Plugging in the known values:
(m_water + m_calorimeter) * 4.18 * (T_final - T_initial_water) = m_copper * 0.39 * (T_initial_copper - T_final)
Here, T_initial_water = 30°C (initial temperature of water)
T_initial_copper = 200°C (initial temperature of copper ball)
We can rearrange the equation to solve for T_final (final temperature):
((m_water + m_calorimeter) * 4.18 * T_final) - ((m_water + m_calorimeter) * 4.18 * T_initial_water) = (m_copper * 0.39 * T_initial_copper) - (m_copper * 0.39 * T_final)
((m_water + m_calorimeter) * 4.18 * T_final) + (m_copper * 0.39 * T_final) = ((m_copper * 0.39 * T_initial_copper) - ((m_water + m_calorimeter) * 4.18 * T_initial_water))
Simplifying the equation further:
((m_water + m_calorimeter) * 4.18 + (m_copper * 0.39)) * T_final = (m_copper * 0.39 * T_initial_copper) - ((m_water + m_calorimeter) * 4.18 * T_initial_water)
Now, we can plug in the values to calculate T_final.
m_water = 50g
m_calorimeter = 60g
m_copper = 20g
T_initial_water = 30°C
T_initial_copper = 200°C
((50 + 60) * 4.18 + (20 * 0.39)) * T_final = (20 * 0.39 * 200) - ((50 + 60) * 4.18 * 30)
Simplifying the equation further:
((110) * 4.18 + (7.8)) * T_final = (7.8 * 200) - ((110) * 4.18 * 30)
(459.8 + 7.8) * T_final = 1564 - (1414.8)
Combining like terms:
467.6 * T_final = 149.2
Dividing by 467.6:
T_final = 0.319 °C
Therefore, the final temperature of the mixture is approximately 0.319°C.
To find the final temperature of the mixture, we can use the principle of conservation of energy. The heat gained by the water and calorimeter is equal to the heat lost by the copper ball.
First, let's calculate the heat gained by the water and calorimeter:
Qwater + Qcalorimeter = mwater * Cwater * ΔTwater + mcalorimeter * Ccalorimeter * ΔTcalorimeter
Where:
Qwater is the heat gained by the water
Qcalorimeter is the heat gained by the calorimeter
mwater is the mass of water
Cwater is the specific heat capacity of water (assumed to be 4.18 J/g°C)
ΔTwater is the change in temperature of the water (final temperature - initial temperature)
mcalorimeter is the mass of the calorimeter
Ccalorimeter is the specific heat capacity of the calorimeter (assumed to be the same as copper, 0.39 J/g°C)
ΔTcalorimeter is the change in temperature of the calorimeter (final temperature - initial temperature)
The initial temperature of the water and calorimeter is 30°C, and we need to find the final temperature, so we'll use T as the final temperature.
Qwater + Qcalorimeter = (50g)(4.18 J/g°C)(T - 30°C) + (60g)(0.39 J/g°C)(T - 30°C)
Next, let's calculate the heat lost by the copper ball:
Qcopper = mcopper * Ccopper * ΔTcopper
Where:
Qcopper is the heat lost by the copper ball
mcopper is the mass of the copper ball
Ccopper is the specific heat capacity of copper (assumed to be 0.39 J/g°C, same as the calorimeter)
ΔTcopper is the change in temperature of the copper ball (final temperature - initial temperature)
The initial temperature of the copper ball is 200°C, and we need to find the final temperature.
Qcopper = (20g)(0.39 J/g°C)(T - 200°C)
Since energy is conserved, the heat lost by the copper ball is equal to the heat gained by the water and calorimeter:
Qcopper = Qwater + Qcalorimeter
Substituting the previously calculated values:
(20g)(0.39 J/g°C)(T - 200°C) = (50g)(4.18 J/g°C)(T - 30°C) + (60g)(0.39 J/g°C)(T - 30°C)
Now, solve the equation to find the final temperature.
20g(0.39 J/g°C)(T - 200°C) = 50g(4.18 J/g°C)(T - 30°C) + 60g(0.39 J/g°C)(T - 30°C)
Simplifying the equation:
7.8(T - 200) = 209(T - 30) + 23.4(T - 30)
Distribute and combine like terms:
7.8T - 1560 = 209T - 6270 + 23.4T - 702
Combine terms:
7.8T - 1560 = 232.4T - 6972
Move all the T terms to one side and all the constant terms to the other side:
7.8T - 232.4T = -6972 + 1560
Combine like terms:
-224.6T = -5412
Divide by -224.6 to solve for T:
T = -5412 / -224.6
T ≈ 24.11°C
Therefore, the final temperature of the mixture is approximately 24.11°C.