A 100.0 g sample of water at 27.0°C is poured into a 77.4 g sample of water at 89.0°C. What will be the final temperature of the water?
To find the final temperature of the water, we can use the principle of conservation of energy.
The formula to calculate the final temperature is:
\(m_1c_1T_1 + m_2c_2T_2 = m_fcf T_f\)
where:
\(m_1\) and \(T_1\) are the mass and initial temperature of the first sample of water (27.0°C)
\(m_2\) and \(T_2\) are the mass and initial temperature of the second sample of water (89.0°C)
\(m_f\) is the total mass of the water after pouring
\(c\) is the specific heat capacity of water (4.18 J/g·°C)
\(T_f\) is the final temperature of the water.
Let's plug in the given values into the formula and solve for \(T_f\).
To find the final temperature of the water, we can use the principle of energy conservation, specifically the equation for heat transfer. The heat gained by the cooler water will be equal to the heat lost by the hotter water.
The equation used for heat transfer is:
Q = m * c * ΔT
Where:
- Q is the heat transferred (in joules)
- m is the mass of the substance (in grams)
- c is the specific heat capacity of the substance (in J/g°C)
- ΔT is the change in temperature (in °C)
First, let's find the heat lost by the hotter water. We can calculate this using the equation stated above:
Q_hot = m_hot * c_water * ΔT_hot
Where:
- Q_hot is the heat lost by the hotter water (in joules)
- m_hot is the mass of the hotter water (in grams)
- c_water is the specific heat capacity of water (approximately 4.18 J/g°C)
- ΔT_hot is the change in temperature of the hotter water (initial temperature - final temperature)
Substituting the given values:
m_hot = 77.4 g
c_water = 4.18 J/g°C
ΔT_hot = 89.0°C - T_final (T_final is the final temperature we want to determine)
Q_hot = 77.4 g * 4.18 J/g°C * (89.0°C - T_final)
Now let's find the heat gained by the cooler water. Again, using the equation for heat transfer:
Q_cool = m_cool * c_water * ΔT_cool
Where:
- Q_cool is the heat gained by the cooler water (in joules)
- m_cool is the mass of the cooler water (in grams)
- ΔT_cool is the change in temperature of the cooler water (final temperature - initial temperature)
Substituting the given values:
m_cool = 100.0 g
c_water = 4.18 J/g°C
ΔT_cool = T_final - 27.0°C
Q_cool = 100.0 g * 4.18 J/g°C * (T_final - 27.0°C)
According to the principle of energy conservation, Q_hot = Q_cool. Therefore, we can equate the equations and solve for T_final:
77.4 g * 4.18 J/g°C * (89.0°C - T_final) = 100.0 g * 4.18 J/g°C * (T_final - 27.0°C)
Solve this equation for T_final to find the final temperature of the water.
heat gained by cool H2O + heat lost by warm H2O = 0
[mass cool H2O x specific heat H2O x (Tfinal-Tinitial)[ + [mass warm H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
Substitute and solve for Tf.
M= 100.0-g
C=2.06J/gC
Ti=116
Tf=75.0