Question

Question #8af18

Answer 1

`18.5^oC`

Explanation:

You need the specific heat of iron in consistent units (i.e. using calories rather than joules). You can find this in the public domain, and it is 0.108 `(cal)/(g.^oC)`.

Then take the relationship `E = m*c*theta`, where E is the energy, m is the mass of substance, c is the specific heat and `theta` is the temperature change.

Rearranging for `theta` we have `theta = E/(m*c)` = `20/(10*0.108) = 18.5^oC`

This, however, assumes no losses to the environment.

You can find specific heats here: www.engineeringtoolbox.com/specific-heat-metals-d_152.html

Note that they are listed in units of `(kcal)/(kg.^oC)`, however, these are numerically equivalent to `(cal)/(g.^oC)` as 1 kcal = 1000 cal, and 1 kg = 1000 g..