A hot lump of 46.9 g of copper at an initial temperature of 54.5 °C is placed in 50.0 mL of H2O initially at 25.0 °C and allowed to reach thermal equilibrium. What is the final temperature of the copper and water given that the specific heat of copper is 0.385 J/(g·°C)? Assume no heat is lost to surroundings.
Plumpy. This only works if you treat each side as absolute values.
The better way to work this is to realize the SUM of the heats gained is zero (one gains, one loses).
mc ΔT (copper) + mc ΔT (water) =0
That way, the math works perfectly.
mc ΔT (copper) = mc ΔT (water)
(46.9 g)(0.385 J/(g.°C))(Tf °C - 54.5 °C) = (50.0 g)(4.18 J/(g.°C))(Tf - 25.0 °C)
Tf = _____ °C
To find the final temperature of the copper and water mixture, we can use the equation for heat transfer:
q = m * c * ΔT
Where:
q = heat transferred (in joules)
m = mass of the substance (in grams)
c = specific heat capacity (in J/(g·°C))
ΔT = change in temperature (in °C)
First, let's find the heat transferred in the system:
q_copper = m_copper * c_copper * ΔT_copper
q_water = m_water * c_water * ΔT_water
Since the system is in thermal equilibrium, the heat gained by the water is equal to the heat lost by the copper:
q_water = -q_copper
Now let's calculate the values for the equation:
m_copper = 46.9 g
c_copper = 0.385 J/(g·°C)
ΔT_copper = Tf_copper - Ti_copper
m_water = 50.0 mL of H2O
c_water = 4.18 J/(g·°C) (specific heat capacity of water at room temperature)
ΔT_water = Tf_water - Ti_water
Assuming the final temperature is Tf for the system, we have:
q_water = -q_copper
m_water * c_water * ΔT_water = -m_copper * c_copper * ΔT_copper
Now let's rearrange the equation to solve for Tf_water:
m_water * c_water * ΔT_water = -m_copper * c_copper * (Tf_copper - Ti_copper)
Substituting the given values:
50.0 mL of H2O = 50.0 g of H2O (since 1 g of water occupies 1 mL)
c_water = 4.18 J/(g·°C)
ΔT_water = Tf_water - 25.0 °C
m_copper = 46.9 g
c_copper = 0.385 J/(g·°C)
ΔT_copper = Tf_copper - 54.5 °C
Ti_copper = 54.5 °C
50.0 g * 4.18 J/(g·°C) * (Tf_water - 25.0 °C) = -46.9 g * 0.385 J/(g·°C) * (Tf_copper - 54.5 °C)
Now let's solve for Tf_water:
209 * (Tf_water - 25.0 °C) = -18.0185 * (Tf_copper - 54.5 °C)
209Tf_water - 209 * 25.0 °C = -18.0185Tf_copper + 18.0185 * 54.5 °C
209Tf_water + 18.0185Tf_copper = 18.0185 * 54.5 °C + 209 * 25.0 °C
Now we need to find the value of Tf_copper in terms of Tf_water. We can use the fact that the heat gained by water equals the heat lost by copper:
m_copper * c_copper * ΔT_copper = -m_water * c_water * ΔT_water
46.9 g * 0.385 J/(g·°C) * (Tf_copper - 54.5 °C) = -50.0 g * 4.18 J/(g·°C) * (Tf_water - 25.0 °C)
18.0185Tf_copper - 18.0185 * 54.5 °C = 209Tf_water - 209 * 25.0 °C
Now we have two equations:
209Tf_water + 18.0185Tf_copper = 18.0185 * 54.5 °C + 209 * 25.0 °C
18.0185Tf_copper + 18.0185 * 54.5 °C = 209Tf_water - 209 * 25.0 °C
Now let's solve these equations using a system of linear equations.
To find the final temperature of the copper and water mixture, we can use the principle of conservation of energy. The heat gained by the water is equal to the heat lost by the copper. The equation to calculate heat transfer is:
q = mcΔT
Where:
q = heat transfer (in joules)
m = mass (in grams)
c = specific heat capacity (in J/(g·°C))
ΔT = change in temperature (in °C)
First, let's calculate the heat gained by the water:
m_water = 50.0 mL = 50.0 g (since the density of water is approximately 1 g/mL)
ΔT_water = Tf_water - Ti_water
Since we want to find the final temperature of the water, we can rearrange the formula to solve for Tf_water:
Tf_water = ΔT_water + Ti_water
To find ΔT_water, we need to know the specific heat capacity of water, which is approximately 4.18 J/(g·°C). Substitute the values into the equation:
ΔT_water = q_water / (m_water * c_water)
ΔT_water = q_copper / (m_water * c_water) [Since q_water = -q_copper according to the principle of conservation of energy]
Now, let's calculate the heat lost by the copper:
m_copper = 46.9 g
Ti_copper = 54.5 °C
c_copper = 0.385 J/(g·°C)
ΔT_copper = Tf_copper - Ti_copper
Since we want to find the final temperature of the copper, we can rearrange the formula to solve for Tf_copper:
Tf_copper = ΔT_copper + Ti_copper
Given that heat is transferred from copper to water, the heat lost by copper is equal to the heat gained by water. So, we can calculate the heat lost by the copper using the equation:
q_copper = q_water = m_water * c_water * ΔT_water
Now we can substitute this value into the equation:
ΔT_copper = q_copper / (m_copper * c_copper)
Finally, substitute the values into the equation for the final temperatures:
Tf_water = ΔT_water + Ti_water
Tf_copper = ΔT_copper + Ti_copper
Calculate these values to find the final temperature of the water and copper.