Question

Two copper blocks, each of mass 2.13 kg, initially have different temperatures,t1 = 19° C and t2 = 33° C. The blocks are placed in contact with each other and come to thermal equilibrium. No heat is lost to the surroundings.

Find the final temperature of the blocks.
______ °C

Find the heat transferred between them.
_____ J

(b) Find the entropy change of each block during the time interval in which the first joule of heat flows.

ΔS1	=  J/K
ΔS2	=  J/K

(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.

ΔS1	=  J/K
ΔS2	=  J/K

Answer 1

A part)we can use the heat formula, which is Q=mCΔT, where C is the specific heat of the material, which is Copper in this case. ΔT for each object, however, is same is magnitude, but it's different in direction. For the first object, the temperature increases which ΔT is positive since the final temperature is creater than initial. On the other hand, the ΔT for second object is negative.

Observing the formula mCΔT we know that

mCΔT=mC(tf-t1)=mC(tf-t2). But since tf-t2 is negative,

mCΔT=mC(tf-t1)=-mC(tf-t2). Since both bodies have the same mass, and C is same for both objects, the equation boils down to,

tf-t1=-(tf-t2)=t2-tf

2*tf=t1+t2,

tf=(t1+t2)/2.

By intuition, we know that the final temperature is half of t1 and t2, which is also proven mathematically.

tf=(19+33)/2=26 degrees Celsius.

For the heat transferred, since there is no loss of heat, heat transferred can be found by either of the temperature. Let's choose t1, (for specific heat of copper 0.385J/g/K=385J/kg/K)

Q=m*C*(tf-t1)=2.13kg*385J/kg/K*(26-19)K (even though the difference is from Celsius, we know the difference is same for Kelvin)

Q=5740.35 J

for part b)

Given the formula for entropy

, the entropy of the first Joule of heat transferred can be estimated.

For the first joule, we can simply use S=Q/T, where Q is 1J, and T can be used as the initial temperature, since the temperature change on the first Joule is very miniscule (to 0.0001 degrees Celsius).

For the first block, more heat is added since the temperature of block 1 is less than block 2.

S1=1J/(19+273)K=0.00342465J/K

For the second block, heat is taken out from the block since the temperature of the second block is higher than the first one,

S2=-1J/(33+273)K=0.00326797 J/K

for part c),

we can use the formula from part b), we can approximate S=Q/T by making T as the average temperature during the process.

We know the change in heat from part a) except that the heat of block 1 after the process is positive and negative for block 2.

The average temperature for block 1's process is (19+273+33+273)/2K=299 K

   for block 2's process is (26+273+33+273)/2K=302.5 K

for the first block,

S1=5740.35 J/299 K=18.976J/K

S2=-5740.35J/302.5K=-18.976 J/K

I hope help you !!