A 110.-g sample of copper (specific heat capacity = 0.20 J/°C · g) is heated to 82.4°C and then placed in a container of water at 22.3°C. The final temperature of the water and copper is 23.5°C. What is the mass of the water in the container, assuming that all the heat lost by the copper is gained by the water?
[nass Cu x specific heat Cu x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
Solve for mass H2O.
To find the mass of the water, we can use the principle of heat transfer, which states that the heat lost by one object is equal to the heat gained by another object when there is no heat exchanged with the surroundings.
First, let's calculate the heat lost by the copper:
Q_copper = m_copper * c_copper * ΔT_copper
Where:
Q_copper represents the heat lost by the copper
m_copper represents the mass of the copper
c_copper represents the specific heat capacity of copper
ΔT_copper represents the change in temperature of the copper
Given:
m_copper = 110.0 g
c_copper = 0.20 J/°C · g
ΔT_copper = 82.4°C - 23.5°C = 58.9°C
Substituting the given values:
Q_copper = 110.0 g * 0.20 J/°C · g * 58.9°C
Now, we know that the heat lost by the copper is equal to the heat gained by the water:
Q_copper = Q_water
We can use the equation for heat transfer in water:
Q_water = m_water * c_water * ΔT_water
Where:
Q_water represents the heat gained by the water
m_water represents the mass of the water
c_water represents the specific heat capacity of water
ΔT_water represents the change in temperature of the water
Given:
c_water = 4.18 J/°C · g (specific heat capacity of water)
Since the final temperature is the same for both copper and water, ΔT_water = 23.5°C - 22.3°C = 1.2°C
Substituting the given values:
Q_water = m_water * 4.18 J/°C · g * 1.2°C
Since Q_copper = Q_water, we can equate the two equations:
110.0 g * 0.20 J/°C · g * 58.9°C = m_water * 4.18 J/°C · g * 1.2°C
Now, we can solve for m_water:
110.0 g * 0.20 J/°C · g * 58.9°C = m_water * 4.18 J/°C · g * 1.2°C
Simplifying:
1291.2 J = m_water * 5.016 J
Dividing both sides by 5.016 J:
m_water = 1291.2 J / 5.016 J
m_water = 257.7 g
Therefore, the mass of the water in the container is approximately 257.7 grams.
To find the mass of the water, we can use the equation for heat transfer:
Q = mcΔT
Where:
Q = heat transferred (in joules)
m = mass of the substance (in grams)
c = specific heat capacity of the substance (in J/°C · g)
ΔT = change in temperature (in °C)
First, let's calculate the heat lost by the copper.
Qcopper = mcopperΔTcopper
Given:
mcopper = 110 g
copper specific heat capacity (ccopper) = 0.20 J/°C · g
ΔTcopper = final temperature of the water and copper - initial temperature of the copper
ΔTcopper = 23.5°C - 82.4°C
Substituting the values into the formula:
Qcopper = (110 g)(0.20 J/°C · g)(-58.9°C)
Next, since all the heat lost by the copper is gained by the water, we can set Qcopper equal to Qwater.
Qwater = mcwaterΔTwater
Given:
ΔTwater = final temperature of the water and copper - initial temperature of the water
ΔTwater = 23.5°C - 22.3°C
Now, we can solve for mcwater:
Qcopper = Qwater
(mcopper)(ccopper)(ΔTcopper) = (mwater)(cwater)(ΔTwater)
Substituting the known values:
(110 g)(0.20 J/°C · g)(-58.9°C) = (mwater)(cwater)(23.5°C - 22.3°C)
Simplifying:
-1293.8 J = (mwater)(cwater)(1.2°C)
Now, we need to know the specific heat capacity of water, which is approximately 4.18 J/g · °C.
Substituting the value:
-1293.8 J = (mwater)(4.18 J/g · °C)(1.2°C)
Dividing both sides by (4.18 J/g · °C)(1.2°C):
-1293.8 J / (4.18 J/g · °C)(1.2°C) = mwater
Calculating:
mwater = -1293.8 J / (4.18 J/g · °C)(1.2°C)
mwater ≈ -259.58 g
The negative sign indicates an error or a mistake made during the calculation. Based on the given information, it is not possible to find the mass of the water in the container using the provided temperatures and final temperature.