Question

Question #ef7b2

Answer 1

`8.41°C`.

Explanation:

First, we need to understand that the heat lost by the water will be the same as the heat gained by the ice.

`q_"water" = q_"ice"`

We know that, if no phase changes take place, heat gained or lost will be `"mass" * "specific heat" * "temperature change"`, or `mCDelta T`.

If we set `x` to be the final temperature, `Delta T` will be `25-x` because the final temperature is less than water's initial temperature.

Therefore, `q_"water"` is:

`q_"water" = "235g" * "4.184 J/g°C" * (25-x)°C`

For ice, the equation is a little more complicated because it involves a phase change. That means we can't just use one `mCDeltaT` to find the total heat gained, because then we would be excluding the heat that was used melt ice.

To solve this problem, let's separate the heat gained by ice into three parts:

1. Ice goes from `–19.0 °C` to `0 °C`.
2. Ice melts to become water at `0 °C`. The heat of fusion of ice is `"333J/g"`, so the total heat gained in this step is `"mass ⋅ 333J/g"`.
3. After becoming water, it gains heat and becomes the final temperature, `x`. The `Delta T` here would be `(x-0)`, or just `x` because the final temperature, `x`, is greater than the initial temperature. The "C" here would be `"4.184J/g°C"`, because the ice is now water! :)

Therefore, `q_"ice"` is:

`q_"ice" = mCDeltaT_1 + "mass ⋅ heat of fusion" + mCDeltaT_2 = ("40g" * "2.09J/g°C" * "19°C") + ("40g ⋅ 333J/g") + ("40g" * "4.184J/g°C" * x°C) = 1588.4J + 13320J + ("167.36J/°C" * x°C)`
`q_"ice" = ("167.36J/°C" * x°C) + 14908.4J`

Remember how `q_"water" = q_"ice"`? We can set those two equations we came up equal to one another with to solve for both of them.

`235* 4.184 * (25-x) = (167.36 * x) + 14908.4`
`983.24 * (25-x) = (167.36 * x )+ 14908.4`
`24581 - 983.24x = 167.36x + 14908.4`
`24581 = 1150.6x + 14908.4`
`9672.6 = 1150.6x`
`x = 8.407`

The least number of significant figures in the question is `3`, so our answer needs to have `3` significant figures too: `8.41°C`.