a 155.0 g piece of copper is dropped into a 250.0 g of water at 28.8 C, what is the final temp of the mixture?
To find the final temperature of the mixture, we can use the principle of conservation of energy.
The amount of heat lost by the copper is equal to the amount of heat gained by the water.
The heat gained or lost can be calculated using the formula:
Q = mcΔT
Where:
Q: heat gained or lost (in joules)
m: mass of the substance (in grams)
c: specific heat capacity of the substance (in J/g°C)
ΔT: change in temperature (in °C)
First, let's calculate the heat lost by the copper.
The specific heat capacity of copper is approximately 0.39 J/g°C.
Mass of copper (m1) = 155.0 g
Initial temperature of copper (T1) = ?
Final temperature of mixture (Tf) = ?
Using the formula, we can calculate the heat lost by the copper:
Q1 = m1c1ΔT1
Next, let's calculate the heat gained by the water.
The specific heat capacity of water is approximately 4.18 J/g°C.
Mass of water (m2) = 250.0 g
Initial temperature of water (T2) = 28.8°C
Final temperature of mixture (Tf) = ?
Using the formula, we can calculate the heat gained by the water:
Q2 = m2c2ΔT2
Since the heat lost by the copper is equal to the heat gained by the water, we can equate the two equations:
Q1 = Q2
(m1c1ΔT1) = (m2c2ΔT2)
Now, we can rearrange the equation to find the final temperature of the mixture (Tf):
Tf = (Q1 + (m2c2ΔT2))/(m1c1)
Substituting the values:
m1 = 155.0 g
c1 = 0.39 J/g°C
ΔT1 = (Tf - T1)
m2 = 250.0 g
c2 = 4.18 J/g°C
ΔT2 = (Tf - T2)
Tf = (m1c1(Tf - T1) + (m2c2(Tf - T2)))/(m1c1 + m2c2)
Now, we can solve this equation to find the final temperature of the mixture.
To find the final temperature of the mixture, we can use the principle of conservation of energy. The heat gained by the water should be equal to the heat lost by the copper, assuming no heat is exchanged with the surroundings.
The equation we can use is:
\(m_1c_1T_1 + m_2c_2T_2 = m_1c_1T_f + m_2c_2T_f\)
Where:
\(m_1\) = mass of copper
\(c_1\) = specific heat capacity of copper
\(T_1\) = initial temperature of copper
\(m_2\) = mass of water
\(c_2\) = specific heat capacity of water
\(T_2\) = initial temperature of water
\(T_f\) = final temperature of the mixture
Let's plug in the values given in the problem:
\(m_1 = 155.0\) g
\(c_1\) (specific heat capacity of copper) is about 0.39 J/g°C
\(T_1\) = initial temperature of copper is not given
\(m_2 = 250.0\) g
\(c_2\) (specific heat capacity of water) is about 4.18 J/g°C
\(T_2 = 28.8\) °C
\(T_f\) = final temperature of the mixture (unknown)
Now, we need to find \(T_1\) using the equation:
\(m_1c_1T_1 + m_2c_2T_2 = m_1c_1T_f + m_2c_2T_f\)
\(155.0 \times 0.39 \times T_1 + 250.0 \times 4.18 \times 28.8 = 155.0 \times 0.39 \times T_f + 250.0 \times 4.18 \times T_f\)
Solving this equation will give us the value of \(T_f\), which is the final temperature of the mixture.