Question

A 1-kg iron block is to be accelerated through a pro- cess that supplies it with 1 kJ of energy. Assuming all this energy appears as kinetic energy, what is the final velocity of the block? b. If the heat capacity of iron is 25.10 J/(mol K) and the molecular weight of iron is 55.85, how large a temperature rise would result from 1 kJ of energy supplied as heat?

Answer 1

Answer:

a) v = 44.72 m/s

b) temperature rise = 2.22 K

Explanation:

A)

Data provided in the question:

Mass of the iron block = 1 kg

Kinetic Energy supplied = 1 KJ = 1000 J

Now,

Kinetic energy = $\frac{1}{2}mv^2$

here,

v is the velocity

thus,

1000 = $\frac{1}{2}\times1\times v^2$

or

v² = 2000

or

v = 44.72 m/s

b)

Heat capacity of iron = 25.10 J/(mol K)

Molecular weight of iron = 55.85

Energy supplied = 1 kJ = 1000 J

Now,

Energy supplied = Mass × C × Change in temperature

Here, C = Heat capacity of iron ÷ Molecular weight of iron

= 25.10 ÷ 55.85

= 0.45 J/(g.K)

thus,

1000 = 1 kg × 0.45 × Change in temperature

or

1000 = 1000 g × 0.45 × Change in temperature

or

Change in temperature i.e temperature rise = 2.22 K