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valina [46]
3 years ago
12

The specific heat capacity of a certain type of cooking oil is 1.75 J/(g⋅∘C). What is the amount of heat exchanged when the temp

erature of 2.76 kg of this oil is cooled from 191∘C to 23 ∘C?
Chemistry
1 answer:
anygoal [31]3 years ago
6 0

Answer:

Q = -811440 J

Explanation:

Given data:

Mass of oil = 2.76 Kg (2.76× 1000 = 2760 g)

Initial temperature = 191 °C

Final temperature = 23°C

Specific heat capacity of oil = 1.75 J/g.°C

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 23°C - 191 °C

ΔT = -168°C

Q = 2760 g ×1.75 J/g.°C  ×-168°C

Q = -811440 J

Negative sign show heat is released.

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Answer:

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Explanation:

Ans: 15.1 grams

Given reaction:

Na2CO3 + Ca(OH)2 → 2NaOH + CaCO3

Mass of Na2CO3 = 20.0 g

Molar mass of Na2CO3 = 105.985 g/mol

# moles of Na2CO3 = 20/105.985 = 0.1887 moles

Based on the reaction stoichiometry: 1 mole of Na2CO3 produces 2 moles of NaOH

# moles of NaOH produced = 0.1887*2 = 0.3774 moles

Molar mass of NaOH = 22.989 + 15.999 + 1.008 = 39.996 g/mol

Mass of NaOH produced = 0.3774*39.996 = 15.09 grams

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