Question

What is the maximum temperature of the water in the insulated container after the copper metal is added? (also steps on how to solve this problem please)

An insulated container is used to hold 41.5 g of water at 35.9 °C. A sample of copper weighing 10.8 g is placed in a dry test tube and heated for 30 minutes in a boiling water bath at 100.0°C. The heated test tube is carefully removed from the water bath with laboratory tongs and inclined so that the copper slides into the water in the insulated container. Given that the specific heat of solid copper is 0.385 J/(g·°C), calculate the maximum temperature of the water in the insulated container after the copper metal is added.

Answer 1

The maximum temperature of the water is `37.4color(white)(l)^oC`.

Explanation:

To solve this problem, we will need to use the specific heat formula:

`Q=mcDeltaT`

Where `Q` is the thermal energy in joules or kilojoules, `c` is the specific heat, and `DeltaT` is the change in temperature, or `T_(fi nal)-T_(i nitial)`.

To start, note that the heat absorbed by the water is equal to the heat released by the copper (assuming no heat loss). So, the following is true:

`Q_(H_2O)=-Q_(Cu)`

Copper releases heat, so its sign is negative, while water absorbs heat, so its sign is positive.

When the copper is added to the water, the water temperature will go up, and the copper temperature will go down as the water absorbs copper's heat.

This means that `T_(fi nal)`, or the final temperatures of both entities will be the same. This temperature is what the question asks for, before the substances start to cool off.

Now, we can plug the known information for both substances into the specific heat formula. We will use the variable `x` for the `T_(fi nal)` that we must solve for.

Note: the specific heat of water is 4.18 J/(g·K).

`Q_(H_2O)=(41.5)(4.18)(x-35.9)`

`Q_(Cu)=(10.8)(0.385)(x-100.0)`

We established previously that `Q_(H_2O)=-Q_(Cu)` and `T_(fi nal)`, or `x` for the two substances are the same. So:

`(41.5)(4.18)(x-35.9)=-(10.8)(0.385)(x-100.0)`

Now, solve for `x` algebraically.

`x=37.4`

This means that the final temperature of the water, or the maximum temperature is `37.4color(white)(l)^oC`.