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When a sample of water (specific heat = 4.18 J/°C) is cooled from 34.0 °C to 28.0°C by placing it in a cold water bath, 85.0 J of energy is released. What is the mass of the water sample?

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Final answer:

The mass of the water sample can be calculated using the formula q = mcΔT, where q is the energy, m is the mass, c is the specific heat, and ΔT is the temperature change. With the given values, the mass of the water sample is found to be 3.39 grams.

Step-by-step explanation:

When a sample of water is cooled from 34.0 °C to 28.0 °C, and releases 85.0 J of energy, we can calculate the mass of the water sample using the specific heat capacity formula q = mcΔT, where q represents the energy in joules, m is the mass in grams, c is the specific heat of water in J/g°C, and ΔT is the change in temperature.

The specific heat of water is given as 4.18 J/g°C. Let's calculate the mass (m) of the water sample:

q = mcΔT → m = q / (cΔT)

We know that q = 85.0 J, c = 4.18 J/g°C, and the temperature change (ΔT) is 34.0 °C – 28.0 °C = 6.0 °C.

Plugging the values into the formula we get:

m = 85.0 J / (4.18 J/g°C × 6.0 °C)

m = 85.0 J / (25.08 J/g)

m = 3.39 g

Therefore, the mass of the water sample is 3.39 grams.

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