Question

Question #b10d9

Answer 1

The final temperature of the mixture will be `"22"^@"C"`.

So, you know that you mix two substances of equal volumes but of different densities, which means that you're mixing two substances of different masses.

If you assume that no temperature is lost to the outside of the beaker, the heat gained by the colder substance, in your case ethanol, will be equal to the heat lost by your warmer substance, in your case water.

`q_("gained") = -q_("lost")`

To solve for the final temperature you need the specific heat of water and that of ethanol, which are

`c_("water") = 4.19"J"/("g" * ^@"C")`

`c_("ethanol") = 2.3"J"/("g" * ^@"C")`

So, mathematically, the equivalence between heat gained and heat lost is written as

`m_("ethanol") * c_("ethanol") * DeltaT_("ethanol") = -m_("water") * c_("water") * DeltaT_("water")` (1)

Determine the masses by using the given volumes and densities

`m_("ethanol") = rho * V = "0.789 g/mL" * "55.0 mL" = "43.4 g"`

`m_("water") = rho * V = "1.00 g/mL" * "55.0 mL" = "55.0 g"`

Now plug your data into the equation (1)

`"43.3"cancel("g") * 2.3cancel("J")/(cancel("g") * ^@cancel("C")) * (color(red)(T_("final")) - 7.0)^@cancel("C") = "55.0"cancel("g") * 4.19cancel("J")/(cancel("g") * ^@cancel("C")) * (28.4 - color(red)(T_("final")))^@cancel("C")`

SIDE NOTE Notice I've removed the minus sign by writing `28.4 - T_("final")` instead of `T_("final") - 28.4`.

Solving this equation for `T_("final")` will give you

`99.82 * color(red)(T_("final")) - 698.74 = 6544.78 - 230.45 * color(red)(T_("final"))`

`color(red)(T_("final")) = 7243.5/330.27 = 21.93^@"C"`

Rounded to two sig figs, the number of sig figs given for `7.0^@C`, the answer will be

`color(red)(T_("final")) = color(red)(22^@"C")`

SIDE NOTE Notice that ethanol's increase in temperature was greater than water's drop in temperature - this is due to the difference between their respective specific heats, i.e. to the difference in energy required to heat one gram of that substance by 1 degree Celsius.