Question #e51aa

The two poles would come to have the same temperature the fastest if the difference between the final temperature and their respective initial temperatures is the same for both of them (in absolute terms), and if they have the same mass.

So, if the two poles are made from the same material, that means they will have the same specific heat. Specific heat is defined as the amount of energy it takes to heat 1 gram of a substance by 1 degree Celsius.

If you start with one cold pole and one warm pole, and you allow them to reach, say, room temperature, this is what would happen:

The heat the cold pole would have to absorb to reach room temperature would be

`q_("cold pole") = m_("cold") * c * DeltaT_("cold")`, where

`m_("cold")` - the cold pole's mass;
`c` - its specific heat;
`DeltaT_("cold")` - the difference between its final temperature and its initial temperature.

The heat the warm pole would have to give off in order to reach room temperature would have to be

`q_("hot pole") = m_("hot") * c * DeltaT_("hot")`

If the two poles give off and absorb the same amount of heat, then they will reach room temperature at the same time. So,

`q_("cold pole") = |-q_("hot pole")|`

(The negative sign on `q_("hot pole")` signifies that its initial temperature was higher than room temperature, therefore `DeltaT_("hot")` will be negative).

`m_("cold") * c * DeltaT_("cold") = m_("hot") * c * |DeltaT_("hot")|`

One way these expressions are equal is if the two poles have the same mass, `m_("cold") = m_("hot")`, and if their `DeltaT`s are equal - in absolute value.

If you start with the cold pole at `0^@C`, the warm one at `50^@C`, and you leave them to rech room temperature, `25^@C`, you'd get

`DeltaT_("cold") = 25 - 0 = 25^@C`
`DeltaT_("hot") = 25 - 50 = -25^@C`

Since `25 = |-25|`, and assuming the poles have the same mass, the will reach room temperature at the same time i. e absorb and give off the same amount of heat.

If the poles don't have the same mass, you could still get them to reach room temperature at the same time if you have a favorable ratio between their masses and their `DeltaT`s, like this

`m_("cold") * DeltaT_("cold") = m_("hot") * |DeltaT_("hot")|`
`100g * 50 = 200g * |-25|`