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.

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

Explanation:

given information :

Temperature of water () =

density of water (ρ) = 1.00 g/mL

Temperature of coffee  () =

volume of coffee () = 180 mL

Mixed Temperature () =

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/)

ΔT = the temperature change ()

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

=

x x Δ = x x Δ

where

is the mass of water

is the mass of coffee

is the specific heat of water

is the specific heat of coffee

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

= , we can remove it from the equation.

Hence.

x Δ = x Δ

we know that

ρ =

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

ρ x V x Δ = ρ x V x Δ

is the volume of water

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

V x Δ = V x Δ

V = (V x Δ) / Δ

V = V x (

V = 180 mL x

V = 180 mL x

V = 170.27 mL

if you're asking what comes out, it's steam.

a pan containing water is placed in a positio n where it is exposed to a 10 mile per hour wind. the pan contains a uniform depth of 1.27 cm of water, is quite wide, and has a length of 2.44 m in the wind direction. the water in the pan is at a constant temperature of 15.5 °c.