A chemist is asked to determine the specific heat capacity of an unknown mineral. The 149-g sample was heated to 92.7°C and placed into a calorimeter containing 81.4 g of water at 20.0°C. The heat capacity of the calorimeter was 12.8 J/K. The final temperature in the calorimeter was 23.7°C. What is the specific heat capacity (in J/g°C) of the mineral? Enter to 4 decimal places.

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jufsanabriasa jufsanabriasa · Answer 1 · 2020-10-20T02:26:48+02:00

Answer:

The specific heat of the mineral is 0.1272J/g°C

Explanation:

The sample is given energy to the calorimeter and the sample of water.

The energy released for the sample is equal to the energy absorbed for both the calorimeter and the water:

C(Sample)*m*ΔT = C(Calorimeter)*ΔT + C(water)*m*ΔT

Where C is specific heat

m is mass of the sample and water

And ΔT is change in temperature

C(Sample)*149g*(92.7°C-23.7°C) = 12.8J/K*(23.7°C-20.0°C) + 4.184J/g°C*81.4g*(23.7°C-20.0°C)

C(Sample)*10281g°C = 47.36J + 1260.1J

C(Sample) = 0.1272J/g°C

andromache andromache · Answer 2 · 2022-02-24T13:35:20+01:00

The specific heat of the mineral is 0.1272J/g°C

The energy released for the sample should be equivalent to the energy absorbed for both the calorimeter and the water:

So,

C(Sample)*m*ΔT = C(Calorimeter)*ΔT + C(water)*m*ΔT

here C is specific heat

m is mass of the sample and water

And ΔT is change in temperature

Now

C(Sample)*149g*(92.7°C-23.7°C) = 12.8J/K*(23.7°C-20.0°C) + 4.184J/g°C*81.4g*(23.7°C-20.0°C)

C(Sample)*10281g°C = 47.36J + 1260.1J

C(Sample) = 0.1272J/g°C

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