A student drops a 0.33-kg piece of steel at 42 ∘C into a container of water at 22 ∘C. The student also drops a 0.51-kg chunk of lead into the same container at the same time. The temperature of the water remains the same.
Part A : Was the temperature of the lead greater than, less than, or equal to 22 ∘C?
Part B: What was the temperature of the lead?
To find the answer to Part A, we need to understand the concept of heat capacity. Heat capacity is the amount of heat energy required to raise the temperature of a substance by 1 degree Celsius. Different substances have different heat capacities.
In this scenario, both the steel and the lead are dropped into the container of water. The temperature of the water remains the same, indicating that the heat energy from the steel and lead is transferred to the water.
Since the temperature of the water remains the same, this implies that the heat gained by the water from the steel must equal the heat gained by the water from the lead.
We can use the formula for heat transfer:
Q = mcΔT
where Q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
Since the initial temperature of the steel is 42 °C and the final temperature of the water is 22 °C, ΔT for the steel is 22 °C.
We can calculate the heat transfer for the steel as follows:
Q_steel = m_steel * c_steel * ΔT_steel
Substituting the known values into the equation:
Q_steel = 0.33 kg * c_steel * 22 °C
Similarly, we can calculate the heat transfer for the lead as follows:
Q_lead = m_lead * c_lead * ΔT_lead
Since the initial temperature of the water is 22 °C, and the temperature of the water remains the same, ΔT for the lead is 0 °C.
Substituting the known values into the equation:
Q_lead = 0.51 kg * c_lead * 0 °C
Since Q_steel = Q_lead, the heat gained by the water from the steel is equal to the heat gained by the water from the lead.
Now, if the heat energy transferred to the water is the same for both the steel and the lead, and the specific heat capacity of lead is higher than steel, this implies that the initial temperature of the lead must be greater than 22 °C for it to transfer the same amount of heat energy to the water.
Therefore, the temperature of the lead (initially) was greater than 22 °C. This answers Part A.
To find the answer to Part B, we need more information. We need to know the specific heat capacity of lead. Once we have that information, we can use the Q_lead equation we derived earlier to calculate the initial temperature.