Please show the correct steps and formulas with explanations

Please show the correct steps and formulas, with explanations, used to solve each part.

Let\'s say we have two 1 kg blocks of metal. One is aluminum and the other is iron. Initially, both blocks are at 25 degree C. a. How much heat must be added to each block to raise each block\'s temperature to 350 degree C? b. If the two heated blocks are then placed in the same insulated container in which the blocks are in thermal communication with each other, what is the final temperature of each block after equilibrium is reached? List any assumptions you make for this simple analysis. c. Let\'s say that only the aluminum block has been heated to 350 degree C while the iron block was kept at 25 degree C. If these two blocks are then placed in the same insulated container in which the blocks are in thermal communication with each other, what is the final temperature of each block after equilibrium is reached? List any assumptions you make for this simple analysis. d. Now reverse the situation described in part c. above. The aluminum block is kept at 25 degree C while the iron block is heated to 350 degree C prior to placing both blocks in the insulated container. What is the final temperature of each block after equilibrium is reached? List any assumptions you make for this simple analysis.

Solution

a)

Heat needed :

H = mi*Ci*(350-25) + ma*Ca*(350 - 25)

where mi = 1 kg

ma = 1 kg

Ca = 0.91 kJ/kg.K

Ci = 0.45 kJ/kg.K

So, H = 1*0.91*10^3*(350-25) + 1*0.45*10^3*(350-25)

= 4.42*10^5 J

b)

As both the bodies are at same thermodynamic temperature, heat wont flow form one to another.

And thus the final temperature of both will be same 350 deg C

c)

Let the final temperature be T

So, heat lost by aluminium = heat gained by iron

So, ma*Ca*(350 - T) = mi*Ci*(T - 25)

So, 1*0.91*(350-T) = 1*0.45*(T - 25)

So, T = 242.5 deg C

d)

Similarly here,

So, heat lost by iron = heat gained by aluminium

So, ma*Ci*(350 -T) = mi*Ca*(T - 25)

So, 1*0.45*(350-T) = 1*0.91*(T - 25)

So, T = 132.5 deg C