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Nutka1998 [239]
3 years ago
5

what is q if 28.6 g of water is heated from 22.0°c to 78.3°c? the specific heat of water is 4.184 j/g·°c.

Physics
1 answer:
Brrunno [24]3 years ago
5 0

Answer:

6736 J

Explanation:

Knowing that Q = cmΔT:

Q = 4.184 J/(g·°C) · 28.6g · (78.3 °C - 22.0 °C) = 6736 J.

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Calculate the change in entropy as 0.3071 kg of ice at 273.15 K melts. (The latent heat of fusion of water is 333000 J / kg)
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Answer:

374.39 J/K

Explanation:

Entropy: This can be defined as the degree of disorder or randomness of a substance.

The S.I unit of entropy is J/K

ΔS = ΔH/T ..................................... Equation 1

Where ΔS = entropy change, ΔH = Heat change, T = temperature.

ΔH = cm................................... Equation 2

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c = specific latent heat of fusion of water = 333000 J/kg, m = mass of ice = 0.3071 kg.

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