what temperature will the water reach when 10.1 g cao is dropped into a coffee cup containing 157 g h2o at 18.0°c if the followi
1 answer:
By using the specific heat capacity of water, the temperature of the water in the cup is found to reach 38.5℃ assuming the cup absorbs negligible heat as it is a perfect insulator.
The specific heat capacity of a substance is defined as the amount of heat (J) absorbed per unit mass (kg) when its temperature rises 1 K (or 1 °C).
The specific heat capacity of a solid is determined by its mass, temperature change, and solid nature.
Specific heat capacity is useful in calculating processing temperatures and the amount of heat required for processing, as well as in distinguishing between two polymeric composite materials.
No. of moles of CaO available = 10.1 / 56 = 0.18 mol
Since 1 mol reactant supplies 64.8 kJ, thus 0.18 mol gives = 11.69 kJ
Therefore, we have 11.69 kJ heat available
Now,
ΔH = mCΔT
where,
ΔH = heat supplied
C = specific heat
ΔT = rise in temperature
Therefore,
ΔT = ΔH/mC = 17.8
Thus,
Final temperature = 18 + 17.8 = 35.8℃
Result:
Hence, (B) 35.8℃ is the temperature of the water in the cup.
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Answer:
2.6 atm
Explanation:
At constant temperature, solubility of gas increases as pressure increases, Hence, they varies directly proportional.
i.e S ∝ P
where:
S₁ and P₁ are the initial solubility and pressure of the gas
S₂ and P₂ are the final solubility and pressure of the gas
Making P₂ the subject of the formula from the above equation; we have:
where; it is given from the question that:
P₁ = 1.0 atm
S₁ = 0.36 g/L
S₂ = 9.5 g/L
Replacing the values into the above equation; we have:
P₂ = 2.6 atm
∴ The pressure needed to produce an aqueous solution containing 9.5 g/L of the same gas at 0°C = 2.6 atm