A 1.1 kg block of iron at 26 C is rapidly heated by a torch such that 15 kJ is t

Question

A 1.1 kg block of iron at 26 C is rapidly heated by a torch such that 15 kJ is transferred to it. What temperature would the block of iron reach assuming the complete transfer of heat and no loss to the surroundings? If the same amount of heat was quickly transferred to a 860 g pellet of copper at 37 C, what temperature would it reach before losing heat to the surroundings? qcs, Fe(s)cs, Cu(s)===mcsT0.450 J/gC0.385 J/gC Express the final temperatures of the iron and copper in C to two significant figures separated by a comma.

Explanation / Answer

Solution:- mass of Iron block, m is given = 1.1 kg = 1100 g

Initial temperature of Iron block = 26 0C

energy gained by Iron block, q = 15 kJ = 15000 J

specific heat capacity, cs of Iron = 0.450 J/(g. 0C)

specific heat capacity, cs of Copper = 0.385 J/(g. 0C)

we know that, q = m cs delta T

where delta T is change in temperature and it is = final temperature - initial temperature

for Iron, 15000 J = 1100 g x 0.450 J/(g. 0C) x delta T

delta T = 15000 J/[1100 g x 0.450 J/(g. 0C)]

delta T = 30 0C

30 0C = final temp - 26 0C

final temp = 30 0C + 26 0C = 56 0C

so, final temperature of Iron block is 56 0C.

Now we could do the calculations for copper in the same way.

15000 J = 860 g x 0.385 J/(g. 0C) x delta T

delta T = 15000 J/[860 g x 0.385 J/(g. 0C)

delta T = 45 0C

45 0C = final temp - 37 0C

final temp = 45 0C + 37 0C = 82 0C

so, the final temperature of copper is 82 0C.

Final temperatures of Fe and Cu are 56 0C, 82 0C

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