How many milliliters of water at 25.0°C with a density of 0.997 g/mL must be mixed with 163 mL of coffee at 97.9°C so that the resulting combination will have a temperature of 54.0°C? Assume that coffee and water have the same density and the same specific heat (4.18 J/g·°C) across the temperature range.

248 mL

Explanation:

According to the law of conservation of energy, the sum of the heat absorbed by water (Qw) and the heat released by the coffee (Qc) is zero.

Qw + Qc = 0

Qw = - Qc [1]

We can calculate each heat using the following expression.

Q = c * m * ΔT

where,

c: specific heat m: mass ΔT: change in the temperature

163 mL of coffee with a density of 0.997 g/mL have a mass of:

163 mL * 0.997 g/mL = 163 g

From [1]

Qw = - Qc

cw * mw * ΔTw = - cc * mc * ΔTc

mw * ΔTw = - mc * ΔTc

mw * (54.0°C-25.0°C) = - 163 g * (54.0°C-97.9°C)

mw * 29.0°C = 163 g * 43.9°C

mw = 247 g

The volume corresponding to 247 g of water is:

247 g * (1 mL/0.997 g) = 248 mL