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?
heat out of steel = heat into lead
since the steel cooled and the lead warmed to reach 22 deg the lead was colder than 22 deg
.33 * Csteel * (42 -22) = .51 * Clead * (22- Tlead)
Part A: Well, the temperature of the water remained the same at 22 ∘C, so we can conclude that the temperature of the lead was also 22 ∘C. It didn't get hotter or colder, just like my sense of fashion.
Part B: So, the temperature of the lead was 22 ∘C, just like a perfectly chilled glass of lemonade on a hot summer day. Refreshing!
To determine the temperature of the lead, we can use the principle of conservation of energy and the specific heat capacity formula. The equation used is:
Q = mcΔT
where Q represents the heat energy transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
Since the temperature of the water remains the same, we can assume that all the energy transferred to the lead is equal to the energy transferred to the water.
For Part A:
Since the temperature of the water remains the same, the energy transferred to the water is equal to zero (Qwater = 0). Since the energy transferred to the lead must be equal to zero as well, this means that the temperature of the lead must be equal to the initial temperature of the water, which is 22 ∘C.
Therefore, the temperature of the lead is equal to 22 ∘C, which means the temperature of the lead is equal to the temperature of the water.
For Part B:
Using the equation Q = mcΔT, we can solve for ΔT:
(mlead * clead * ΔTlead) = -(mwater * cwater * ΔTwater)
Since the mass of the water is not specified and the temperature remains the same, we can assume the energy transferred to the water is equal to the energy transferred to the lead.
(0.33 kg * csteel * ΔTlead) = -(0.51 kg * clead * (22 ∘C - 42 ∘C))
Since steel and lead have different specific heat capacities, we need to know the specific heat capacity of the steel (csteel) and lead (clead) in order to solve for the change in temperature (ΔTlead).
Without that information, we cannot calculate the exact temperature of the lead.
To answer this question, we need to understand the concept of thermal equilibrium and heat transfer.
Part A: To determine whether the temperature of the lead was greater than, less than, or equal to 22 ∘C, we need to consider the heat transfer between the lead and the water. If the lead had a higher temperature than the water, it would transfer heat to the water until both reached a common temperature. If the lead had a lower temperature, the water would transfer heat to the lead. If the temperature of the lead was equal to the water, there would be no heat transfer.
Part B: To find the temperature of the lead, we can use the heat transfer equation, which states:
Q = mcΔT
Where:
Q is the heat transferred (in joules),
m is the mass of the substance (in kilograms),
c is the specific heat capacity of the substance (in joules per kilogram per degree Celsius),
and ΔT is the change in temperature (in degrees Celsius).
Since the temperature of the water remains the same, the heat transferred to the water by the lead is equal to the heat transferred from the steel. We can equate the two expressions to find the temperature of the lead.
Qlead = Qsteel
mlead * clead * ΔT_lead = msteel * csteel * ΔT_steel
Given the mass of the lead (0.51 kg) and the steel (0.33 kg), and assuming the specific heat capacity of lead and steel are constant, we can solve for the temperature of the lead.
Please provide the specific heat capacities of lead and steel.