Calculate the specific heat capacity of substance x from the following data. A container made of the substance x has a mass of 3.6kg and contains 14kg of water. A 1.8kg-piece of the substance x initially at the temperature of 180°C is dropped into the water. The container and the water initially have a temperature of 16.0°C and the final temperature of the system is 18.0°C.
the sum of heats gained is zero
3.6*cx*(18-16)+ 14*cw*(18-16)+1.8*cx*(18-180)=0
solve for cx
411
To calculate the specific heat capacity of substance x, we can use the formula:
Q = mcΔT
Where:
Q is the heat transferred
m is the mass
c is the specific heat capacity
ΔT is the change in temperature
First, let's calculate the heat transferred to the water:
Q_water = mc_waterΔT_water
Where:
m_water = mass of water = 14 kg
c_water = specific heat capacity of water = 4.186 J/g°C (or 4186 J/kg°C)
ΔT_water = change in temperature of the water = final temperature - initial temperature = 18.0°C - 16.0°C = 2.0°C
Now, let's calculate the heat transferred to the container:
Q_container = mc_containerΔT_container
Where:
m_container = mass of container = 3.6 kg
c_container = specific heat capacity of substance x (what we need to find)
ΔT_container = change in temperature of the container = final temperature - initial temperature = 18.0°C - 16.0°C = 2.0°C
Since the heat transferred to the container and the heat transferred to the water come from the same initial source, they are equal. Therefore:
Q_water = Q_container
mc_waterΔT_water = mc_containerΔT_container
Substituting the values we have:
14 kg * 4.186 J/kg°C * 2.0°C = 3.6 kg * c_container * 2.0°C
Simplifying the equation:
117.404 J = 7.2 kg * c_container
Dividing both sides by 7.2 kg:
c_container = 117.404 J / 7.2 kg ≈ 16.28 J/kg°C
Therefore, the specific heat capacity of substance x is approximately 16.28 J/kg°C.
To calculate the specific heat capacity of substance X, we can use the equation:
Q = m x c x ΔT
Where:
Q is the heat transferred
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature
First, let's find the heat transferred from the substance X to the water.
The formula for the heat transferred (Q) can be written as:
Q = Q_substance_X + Q_water
Where Q_substance_X is the heat transferred from the substance X, and Q_water is the heat transferred to the water.
To find Q_substance_X, we can use the equation:
Q_substance_X = m_substance_X x c_substance_X x ΔT_substance_X
Given:
m_substance_X = 1.8 kg (mass of the substance X)
c_substance_X = ? (specific heat capacity of the substance X)
ΔT_substance_X = (18.0°C - 180°C) = -162°C (change in temperature of the substance X)
Since we need to calculate c_substance_X, we can rearrange the equation to solve for c_substance_X:
c_substance_X = Q_substance_X / (m_substance_X x ΔT_substance_X)
Now, let's calculate Q_substance_X:
Q_substance_X = m_substance_X x c_substance_X x ΔT_substance_X
Given:
m_substance_X = 1.8 kg
ΔT_substance_X = -162°C
To convert ΔT from Celsius to Kelvin, we add 273:
ΔT_substance_X = -162 + 273 = 111 K
Now, we can find Q_substance_X:
Q_substance_X = (1.8 kg) x (c_substance_X) x (111 K)
Next, let's find Q_water:
Q_water = m_water x c_water x ΔT_water
Given:
m_water = 14 kg (mass of the water)
c_water = 4.18 kJ/kg°C (specific heat capacity of water, a commonly used value)
ΔT_water = (18.0°C - 16.0°C) = 2.0°C (change in temperature of the water)
Since we already know the values for m_water, c_water, and ΔT_water, we can calculate Q_water:
Q_water = (14 kg) x (4.18 kJ/kg°C) x (2.0°C)
Now, substitute the values of Q_substance_X and Q_water into the initial equation:
Q = Q_substance_X + Q_water
Since we know that the heat transfer is conserved in an isolated system (no heat loss to the surroundings), the heat transferred from the substance X is equal to the heat gained by the water:
Q_substance_X = -Q_water
Therefore,
Q_substance_X = -Q_water
(1.8 kg) x (c_substance_X) x (111 K) = - (14 kg) x (4.18 kJ/kg°C) x (2.0°C)
Simplifying the equation:
(1.8 kg) x (c_substance_X) x (111 K) = - (14 kg) x (4.18 kJ/kg°C) x (2.0°C)
(1.8 kg) x (c_substance_X) x (111 K) = - (14 kg) x (8.36 kJ)
Solve for c_substance_X:
c_substance_X = [(14 kg) x (8.36 kJ)] / [(1.8 kg) x (111 K)]
Evaluate the expression:
c_substance_X = 102.4 kJ/kg°C
Therefore, the specific heat capacity of substance X is 102.4 kJ/kg°C.