Question

A 5.10 g sample of iron is heated from 36.0 to 75.0 C. The amount of energy required is 89.5 J. The specific capacity of this sample of iron is...

A.) 17800 J/g C
B.)0.900 J/g C
C.)11.7 J/g C
D.) 0.450 J/g C
E.) 2.22 J/g C

Answer 1

This can be solved from the equation q = m * C * ΔT Where q is the heat added, m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.
We can rearrange the equation to find the specific heat capacity
q/(m*ΔT) = C
q = 89.5 J m = 5.10 g ΔT = (75.0 °C - 36.0 °C) = 39 °C
(89.5J)/(5.10 g * 39 °C) = 0.450 J/g°C

Answer 2

Answer: Option (D) is the correct answer.

Step-by-step explanation:

The given data is as follows.

mass = 5.10 g,
`T_(1) = 36^(o)C`,

`T_(2) = 75.0^(o)C`, q = 89.5 J

It is known that relation between heat energy and specific heat capacity is as follows.

q =
`mC \Delta T`

Hence, putting the given values into the above formula to calculate specific heat capacity as follows.

q =
`mC \Delta T`

89.5 J =
`5.10 g * C (75 - 36)^(o)C`

89.5 J =
`5.10 g * C * 39^(o)C`

C =
`0.450 J/g ^(o)C`

Thus, we can conclude that the specific capacity of this sample of iron is
`0.450 J/g ^(o)C`.