Question

The specific heat capacity of an unknown liquid is `0.32` `"J"/("kgK")`. The density of the liquid is `0.0321` `"g/mL"`. If a chemist applies `243` `"J"` of heat to `300` `"mL"` of this liquid starting at `27.1 ^@` `C`, what is the final temperature?

Answer is `80000^@"C"`

Thanks in advance

Answer 1

Given,

`C_x = (3.2*10^-4"J")/("g"*"°C")`

`rho_x = (0.0321"g")/"mL"`

`V = 300"mL"`

`T_"i" = 27.1°"C"`

`q = 243"J"` (we're adding heat to the system, so its sign will be positive)

Now, recall,

`q=mC_sDeltaT`, and

`rho = m/V`

`=> m = rho_xV`

Hence,

`q = rho_xVC_x(T_"f" - T_"i")`

`=> T_"f" = q/(rho_xVC_x) + T_"i" approx 8*10^4°"C"`

is the final temperature of the solution, given your data.

Answer 2

80000 C

Explanation:

So we start with the equation q=mc`Delta`t

Since `Delta`T= The final temp. (`T_f`) - The initial temp. (`T_i`)

We're given the initial temp. (`T_i`) which is `27.1 C`
However, we're not given the final temp. (`T_f`) or the `Delta`T

This means we first have to find, `Delta`T first

We do this by taking the equation q=mc`Delta`t and rearranging it so we isolate the `Delta`T

It should now look like
q/mc =`Delta`t

So we have our rearranged equation, but we first need to do some conversions before we plug anything in.

The `300 mL` needs to be in `kg`

Since we are told that `0.0321 g` is equal to `1mL`, first convert the `mL` to `g`

`300cancel(mL)` x `(0.0321g)/(1cancel(mL))` = `9.63g`

Now we have to convert `g` into `kg`
`1000g = 1kg`

`9.63cancel(g)` x `(1 kg)/(1000cancel(g))` = `0.00963kg`

Now that is done, we can plug our values into the equation
q/mc =`Delta`t

Our `q` is our Joules (`243J`)
Our `m` is our mass (`0.00963kg`)
Our `c` is our specific heat capacity ( `(0.32J)/(KgK)`)

So now with terms plugged in, it should look like this

(`243J`) / (`0.00963kg`) (`(0.32J)/(KgK)`) = `Delta`T

Our Joules and kg will cancel out, and leave us with `K`, which is kelvin
(`243cancelJ`) / (`0.00963cancel(kg)`) (`(0.32cancel(J))/(cancel(kg)K)`) = `Delta`T

Now multiply the `0.00963` and `0.32` to get `0.0030816`
`243` / `0.0030816K` = `Delta`T

Now divide the `243` by `0.0030816K`

So `Delta`T= `78,855.14K`

Since our final answer is in `C`, we need to convert `K` to `C`

We do this by subtracting `273`

`78,855.14` -`273` = `78,582.14` `C`

So now our `Delta`T = `78,582.14` `C`

Then we take our equation, `Delta`T= The final temp. (`T_f`) - The initial temp. (`T_i`) and plug our values in

`Delta`T = `78,582.14` `C`
`T_i`=`27.1 C`

`78,582.14` `C` = `T_f` - `27.1 C`

Add `27.1 C` to - `27.1 C` and `78,582.14` `C` to isolate `T_f`

Your final answer should be
`78,609.24` `C` = `T_f`
But with it rounded, it's `80,000` `C`