Two copper blocks, each of mass 2.03 kg, initially have different temperatures,t1 = 15° C and t2 = 30° C. The blocks are placed in contact with each other and come to thermal equilibrium. No heat is lost to the surroundings.
and the question is....
(a) Find the final temperature of the blocks
Find the heat transferred between them.
(b) Find the entropy change of each block during the time interval in which the first joule of heat flows.
Delta S1=
Delta S2=
(c) Estimate the entropy change of each block after it has reached thermal equilibrium. Use each block's average temperature during the process in calculating the estimated values of ΔS.
To determine the final temperature of the copper blocks when they reach thermal equilibrium, we can use the principle of thermal equilibrium, which states that when two objects are in thermal contact, heat will flow from the object with a higher temperature to the object with a lower temperature until they reach the same final temperature.
The heat transferred between two objects can be calculated using the formula:
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 no heat is lost to the surroundings, the heat gained by one block is equal to the heat lost by the other block. Therefore, we equate the heat transferred for both blocks and solve for the final temperature.
For the block initially at 15°C:
Q1 = mcΔT1
Q1 = (2.03 kg)(Specific heat capacity of copper)(Final temperature - 15°C)
For the block initially at 30°C:
Q2 = mcΔT2
Q2 = (2.03 kg)(Specific heat capacity of copper)(Final temperature - 30°C)
Since Q1 = Q2, we can equate the two equations above and solve for the final temperature. Note that the specific heat capacity of copper is approximately 387 J/(kg°C).
However, the specific heat capacity of copper is given in J/(kg°C), but we need to convert the mass from kg to grams and the temperature from °C to Kelvin (K) before plugging in the values.
Once we have the final temperature in Kelvin, we can convert it back to Celsius by subtracting 273.15.
Hope this helps! Let me know if you have any further questions.