Two blocks of different temperatures are brought together.
Block A is a 360.0-kg block is made of Material X which has a temperature-dependent specific heat. Its initial temperature is TA = 200.0¡ã C. Block B is a 1000.0-kg iron block with an initial temperature TB = 27.0¡ã C. The specific heat of Material X in the range of interest is cA = T*0.64 J/(kg K2) and the specific heat of iron is cB = 452.0 J/(kg K).
We bring the two blocks together and wait for them to reach a common equilibrium temperature Tf. If we assume that the two block system does not lose energy to the environment, then what is the change in entropy of the two block system?
0oC = 273.15 K
¦¤S =
erwf
To find the change in entropy of the two-block system, we need to consider the change in entropy for each block individually and then sum them up.
First, let's find the change in entropy for Block A. The change in entropy (ΔS) is given by the equation:
ΔS = ∫(dq / T)
Where dq is a small amount of heat transferred to or from the block and T is the temperature.
Since the specific heat of Material X is temperature-dependent (cA = T*0.64 J/(kg K^2)), we need to use this equation to calculate dq for each small temperature interval and then sum them up.
dq = cA*dT
Where dT is a small change in temperature.
The temperature range for Block A is from its initial temperature (TA) to the equilibrium temperature (Tf). Therefore, the change in entropy for Block A can be calculated as follows:
ΔSA = ∫(cA*dT / T)
Now, let's find the change in entropy for Block B. The specific heat of iron (cB) is constant, so we can directly use the equation:
ΔSB = ∫(cB*dT / T)
Again, the temperature range for Block B is from its initial temperature (TB) to the equilibrium temperature (Tf).
Now, let's find the sum of the changes in entropy for both blocks:
ΔS = ΔSA + ΔSB
You can substitute the specific heat values and integrate the expressions to get the change in entropy for the two-block system. Remember to convert the temperatures to Kelvin before performing any calculations.