Answer:
The number of water that must be mixed in the solution is 170.27 mL
Explanation:
given information :
Temperature of water ([tex]T_{1}[/tex]) = [tex]23^{o}C[/tex]
density of water (ρ) = 1.00 g/mL
Temperature of coffee ([tex]T_{1}[/tex]) = [tex]95^{o}C[/tex]
volume of coffee ([tex]V_{2}[/tex]) = 180 mL
Mixed Temperature ([tex]T_{mix}[/tex]) = [tex]60^{o}C[/tex]
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/[tex]g^{o}C[/tex])
ΔT = the temperature change ([tex]^{o}C[/tex])
in the mixture solution, the heat of the water ([tex]Q_{1}[/tex]) should be the same as the heat of coffee ([tex]Q_{2}[/tex]). Thus,
[tex]Q_{1}[/tex] = [tex]Q_{2}[/tex]
[tex]m_{1}[/tex] x [tex]c_{1}[/tex] x Δ[tex]T_{1}[/tex] = [tex]m_{2}[/tex] x [tex]c_{2}[/tex] x Δ[tex]T_{2}[/tex]
where
[tex]m_{1}[/tex] is the mass of water
[tex]m_{2}[/tex] is the mass of coffee
[tex]c_{1}[/tex] is the specific heat of water
[tex]c_{2}[/tex] is the specific heat of coffee
Assume coffee and water have the same specific heat. So,
[tex]c_{1}[/tex] = [tex]c_{2}[/tex], we can remove it from the equation.
Hence.
[tex]m_{1}[/tex] x Δ[tex]T_{1}[/tex] = [tex]m_{2}[/tex] x Δ[tex]T_{2}[/tex]
we know that
ρ = [tex]\frac{m}{V}[/tex]
m = ρ x V, subtitute it to the equation:
ρ[tex]_{1}[/tex] x V[tex]_{1}[/tex] x Δ[tex]T_{1}[/tex] = ρ[tex]_{2}[/tex] x V[tex]_{2}[/tex] x Δ[tex]T_{2}[/tex]
[tex]V_{1}[/tex] is the volume of water
coffee and water have the same density, so we can remove the formula
V[tex]_{1}[/tex] x Δ[tex]T_{1}[/tex] = V[tex]_{2}[/tex] x Δ[tex]T_{2}[/tex]
V[tex]_{1}[/tex] = (V[tex]_{2}[/tex] x Δ[tex]T_{2}[/tex]) / Δ[tex]T_{1}[/tex]
V[tex]_{1}[/tex] = V[tex]_{2}[/tex] x ([tex]\frac{(T_{2} - T_{mix}) }{(T_{mix} - T_{1})}[/tex]
V[tex]_{1}[/tex] = 180 mL x [tex]\frac{(95-60)^{o}C}{(60-23)^{o}C}[/tex]
V[tex]_{1}[/tex] = 180 mL x [tex]\frac{(35)^{o}C}{(37)^{o}C}[/tex]
V[tex]_{1}[/tex] = 170.27 mL
