Question

18. How many milliliters of water at 23 °C with a density of 1.00 g/mL must be mixed with 180 mL (about 6 oz) of coffee at 95 °C so that the resulting combination will have a temperature of 60 °C? Assume that coffee and water have the same density and the same specific heat.

Answer 1

The number of water that must be mixed in the solution is 170.27 mL

Step-by-step explanation:

given information :

Temperature of water (
`T_(1)`) =
`23^(o)C`

density of water (ρ) = 1.00 g/mL

Temperature of coffee (
`T_(1)`) =
`95^(o)C`

volume of coffee (
`V_(2)`) = 180 mL

Mixed Temperature (
`T_(mix)`) =
`60^(o)C`

to calculate the heat, w use the formula :

Q = m x c x ΔT

where

m = mass of the substance (g)

c = specific heat (J/
`g^(o)C`)

ΔT = the temperature change (
`^(o)C`)

in the mixture solution, the heat of the water (
`Q_(1)`) should be the same as the heat of coffee (
`Q_(2)`). Thus,

`Q_(1)` =
`Q_(2)`

`m_(1)` x
`c_(1)` x Δ
`T_(1)` =
`m_(2)` x
`c_(2)` x Δ
`T_(2)`

where

`m_(1)` is the mass of water

`m_(2)` is the mass of coffee

`c_(1)` is the specific heat of water

`c_(2)` is the specific heat of coffee

Assume coffee and water have the same specific heat. So,

`c_(1)` =
`c_(2)`, we can remove it from the equation.

Hence.

`m_(1)` x Δ
`T_(1)` =
`m_(2)` x Δ
`T_(2)`

we know that

ρ =
`(m)/(V)`

m = ρ x V, subtitute it to the equation:

ρ
`_(1)` x V
`_(1)` x Δ
`T_(1)` = ρ
`_(2)` x V
`_(2)` x Δ
`T_(2)`

`V_(1)` is the volume of water

coffee and water have the same density, so we can remove the formula

V
`_(1)` x Δ
`T_(1)` = V
`_(2)` x Δ
`T_(2)`

V
`_(1)` = (V
`_(2)` x Δ
`T_(2)`) / Δ
`T_(1)`

V
`_(1)` = V
`_(2)` x (
`((T_(2) - T_(mix)) )/((T_(mix) - T_(1)))`

V
`_(1)` = 180 mL x
`((95-60)^(o)C)/((60-23)^(o)C)`

V
`_(1)` = 180 mL x
`((35)^(o)C)/((37)^(o)C)`

V
`_(1)` = 170.27 mL