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2 months ago

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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.

1 answer:

Anarel [2.9K]2 months ago

3 0

Answer:

The amount of water needed to mix with the solution is 170.27 mL

Explanation:

Provided details:

Water temperature ( $T_{1}$ ) = $23^{o}C$

density of water (ρ) = 1.00 g/mL

Coffee temperature ( $T_{1}$ ) = $95^{o}C$

Coffee volume ( $V_{2}$ ) = 180 mL

Mixed Temperature ( $T_{mix}$ ) = $60^{o}C$

To calculate the heat, we apply the formula:

Q = m x c x ΔT

where

m = mass (g)

c = specific heat (J/ $g^{o}C$ )

ΔT = temperature change ( $^{o}C$ )

In this mixed solution, the water's heat ( $Q_{1}$ ) must equal the coffee's heat ( $Q_{2}$ ). 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 water’s mass

is the coffee’s mass $m_{2}$

refers to the specific heat of water

$c_{1}$ and

is the specific heat of coffee $c_{2}$

Assuming the specific heats of both coffee and water are identical, therefore,

$c_{1}$ = $c_{2}$ , can be canceled in the equation.

Thus,.

$m_{1}$ x Δ $T_{1}$ = $m_{2}$ x Δ $T_{2}$

We know that

ρ = $\frac{m}{V}$

m = ρ x V, substitute this in 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

Since coffee and water share the same density, hence, we can eliminate the formula

V $_{1}$ x Δ $T_{1}$ = V $_{2}$ x Δ $T_{2}$

Thus, V $_{1}$ = (V $_{2}$ x Δ $T_{2}$ ) / Δ $T_{1}$ V $_{1}$ = V $_{2}$ x ( $\frac{(T_{2} - T_{mix}) }{(T_{mix} - T_{1})}$

V $_{1}$ = 180 mL x $\frac{(95-60)^{o}C}{(60-23)^{o}C}$

V $_{1}$ = 180 mL x $\frac{(35)^{o}C}{(37)^{o}C}$

V $_{1}$ = 170.27 mL

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