Question

A 1.1 kg block of iron at 38 ∘C is rapidly heated by a torch such that 13 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 890 g pellet of copper at 38 ∘C, what temperature would it reach before losing heat to the surroundings?

Cs, Fe(s)= 0.450 J/g*C
Cs, Cu(s)= 0.385 J/g*C

Answer 1

Answer:- Iron block final temperature is 64 degree C and copper block final temperature is 76 degree C.

Solution:- These problems are based on the formula:

`q=mC_s\Delta T`

where, q is the heat energy, m is mass,
`C_s` is specific heat and
`\Delta T` is the change in temperature.

mass of iron is given as 1.1 kg. We need to convert it to grams:

`1.1kg((1000g)/(1kg))` = 1100 g

q is 13 kJ, let's convert it to J:

`13kJ((1000J)/(1kJ))` = 13000 J

`C_s` of iron is
`(0.450J)/(g.^0C)`

Let's calculate change in temperature for iron.

`\Delta T=(13000J)/(1100g*(0.450J)/(g.^0C))`

`\Delta T=26 ^0C`

Since,
`\Delta T` = final temperature - initial temperature

26 = final temperature - 38

final temperature = 26 + 38 = 64 degree C

So, The final temperature of iron block would be 64 degree C.

Let's do the same calculations for copper block, it's mass is 890 g and specific heat is
`(0.385J)/(g.^0C)` .

`\Delta T=(13000J)/(890g*(0.385J)/(g.^0C))`

`\Delta T=38 ^0C`

Now, calculations for final temperature:

`\Delta T` = final temperature - initial temperature

38 = final temperature - 38

final temperature = 38 + 38 = 76 degree C

So, the final temperature of copper block is 76 degree C.