A 2.41 g lead weight, initially at 10.7 ∘C, is submerged in 8.08 g of water at 51.9 ∘C in an insulated container.
Question:
What is the final temperature of both the weight and the water at thermal equilibrium?
Express the temperature in Celsius to three significant figures.
[mass Pb x specific heat Pb x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
The only unknown here is Tfinal
To find the final temperature of both the lead weight and the water at thermal equilibrium, we can use the principle of energy conservation.
The principle of energy conservation states that the total energy of an isolated system remains constant.
In this case, the system consists of the lead weight and the water.
The energy transfer happens until both the weight and the water reach the same final temperature.
To find this final temperature, we can use the principle that the heat gained by one substance is equal to the heat lost by the other substance.
The heat gained by the lead weight can be calculated using the formula:
Q1 = m1 * c1 * ΔT1
Where:
Q1 = heat gained by the lead weight
m1 = mass of the lead weight (2.41 g)
c1 = specific heat capacity of lead
ΔT1 = change in temperature of the lead weight
Similarly, the heat lost by the water can be calculated using the formula:
Q2 = m2 * c2 * ΔT2
Where:
Q2 = heat lost by the water
m2 = mass of the water (8.08 g)
c2 = specific heat capacity of water
ΔT2 = change in temperature of the water
Since the final temperature is the same for both the weight and the water, we can set Q1 equal to Q2:
Q1 = Q2
m1 * c1 * ΔT1 = m2 * c2 * ΔT2
Now, we need to rearrange the equation to solve for the final temperature.
First, let's calculate ΔT1 and ΔT2:
ΔT1 = final temperature - initial temperature of the lead weight
ΔT2 = final temperature - initial temperature of the water
Now, substitute the values into the equation:
m1 * c1 * (final temperature - initial temperature of the lead weight) = m2 * c2 * (final temperature - initial temperature of the water)
Substitute the given values:
2.41 g * c1 * (final temperature - 10.7 °C) = 8.08 g * c2 * (final temperature - 51.9 °C)
Now, let's solve for the final temperature numerically.
The specific heat capacity of lead (c1) is 0.129 J/g·°C.
The specific heat capacity of water (c2) is 4.18 J/g·°C.
Substituting the values, we have:
2.41 g * 0.129 J/g·°C * (final temperature - 10.7 °C) = 8.08 g * 4.18 J/g·°C * (final temperature - 51.9 °C)
Now we can solve for the final temperature.