A 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C. If the final temperature of water is 42.0 °C, what was the initial temperature of the copper piece? (5 points)

Specific heat of copper = 0.39 J/g °C

Group of answer choices

322 °C

345 °C

356 °C

364 °C

Answers

Answer 1

The initial temperature of the copper piece if a 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C is 345.5°C

How to calculate temperature?

The initial temperature of the copper metal can be calculated using the following formula on calorimetry:

Q = mc∆T

mc∆T (water) = - mc∆T (metal)

Where;

m = massc = specific heat capacity∆T = change in temperature

According to this question, a 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C. If the final temperature of water is 42.0 °C, the initial temperature of the copper is as follows:

400 × 4.18 × (42°C - 24°C) = 240 × 0.39 × (T - 24°C)

30,096 = 93.6T - 2246.4

93.6T = 32342.4

T = 345.5°C

Therefore, the initial temperature of the copper piece if a 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C is 345.5°C.

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Answer 2

Answer:

the correct answer is 364

q = m * c * ΔT

where q is the heat absorbed or released, m is the mass of the substance, c is the specific heat capacity of the substance, and ΔT is the change in temperature.

We can assume that the heat lost by the copper is gained by the water, so:

q(copper) = -q(water)

where the negative sign indicates that the copper loses heat while the water gains heat.

The specific heat capacity of copper is 0.39 J/g °C, so:

q(copper) = 240.0 g * 0.39 J/g °C * (T(copper) - 24.0 °C)

The specific heat capacity of water is 4.18 J/g °C, so:

q(water) = 400.0 g * 4.18 J/g °C * (42.0 °C - T(copper))

Setting q(copper) equal to -q(water), we get:

240.0 g * 0.39 J/g °C * (T(copper) - 24.0 °C) = -400.0 g * 4.18 J/g °C * (T(copper) - 42.0 °C)

Simplifying and solving for T(copper), we get:

T(copper) = [(400.0 g * 4.18 J/g °C * 42.0 °C) + (240.0 g * 0.39 J/g °C * 24.0 °C)] / (240.0 g * 0.39 J/g °C + 400.0 g * 4.18 J/g °C)

T(copper) = 364.1 °C

Therefore, the initial temperature of the copper piece was 364.1 °C.

Hence, the answer is "364 °C".


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