Answer:
a) Q₁ = 58.82 KJ
b) Q₂ = 42.48 KJ
c) Q₃ = 29.22 KJ
d) Q = 130.52 KJ
Explanation:
a)
In order to find the energy absorbed to heat the solid, we will use:
[tex]Q_{1} = mC_{1}\Delta T_{1}[/tex]
where,
Q₁ = Heat absorbed for heating solid = ?
m = mass of solid = 488.3 g = 0.4883 kg
C₁ = Specific Heat Capacity of Solid = 2.96 J/g °C
ΔT₁ = Change in temperature of Solid = Melting Temperature - Initial Temp.
ΔT₁ = 17.6°C - (-23.1°C) = 40.7°C
Therefore,
[tex]Q_{1} = (488.3\ g)(2.96\ J/g\ ^{0}C)(40.7\ ^{0}C)[/tex]
Q₁ = 58.82 KJ
b)
In order to find the absorbed to melt the solid at 17.6°C, we will use:
[tex]Q_{2} = nH_{fus}[/tex]
where,
Q₂ = Heat absorbed for melting solid = ?
H_fus = Heat of Fusion = 8.04 KJ/mol
n = no. of moles = [tex]\frac{m}{MM} = \frac{488.3\ g}{92.41\ g/mol} = 5.28 mol[/tex]
Therefore,
[tex]Q_{2} = (5.28\ mol)(8.04\ KJ/mol)[/tex]
Q₂ = 42.48 KJ
c)
In order to find the energy absorbed to heat the liquid, we will use:
[tex]Q_{3} = m C_{3}\Delta T_{3}[/tex]
where,
Q₃ = Heat absorbed for heating Liquid = ?
m = mass of solid = 488.3 g = 0.4883 kg
C₃ = Specific Heat Capacity of Liquid = 1.75 J/g °C
ΔT₃ = Change in temperature of Liquid = Final Temp. - Melting Temp.
ΔT₃ = 51.8°C - 17.6°C = 34.2°C
Therefore,
[tex]Q_{3} = (488.3\ g)(1.75\ J/g\ ^{0}C)(34.2\ ^{0}C)[/tex]
Q₃ = 29.22 KJ
d)
Total amount of energy absorbed during entire process is:
[tex]Q = Q_{1} + Q_{2} + Q_{3}[/tex]
[tex]Q = 58.82\ KJ + 42.48\ KJ + 29.22\ KJ[/tex]
Q = 130.52 KJ
In order to heat a 488.3 g solid, 58.8 kJ are required. To melt the solid, 42.5 kJ are required. To heat the liquid, 29.2 kJ are required. The total amount of energy absorbed is 130.5 kJ.
Initially, a 488.3 g solid at -23.1 °C is heated up to 17.6 °C (melting point). We can calculate the heat required (Q₁) using the following expression.
[tex]Q_1 = c \times m \times \Delta T = \frac{2.96J}{g.\° C } \times 488.3g \times (17.6\° C-(-23.1\° C)) \times \frac{1kJ}{1000J} = 58.8 kJ[/tex]
where,
At 17.6 °C, we can calculate the heat (Q₂) required to melt the solid using the following expression.
[tex]Q_2 = \Delta H_{fus} \times \frac{m}{MM} = 8.04 kJ/mol \times \frac{488.3 g}{92.41g/mol} = 42.5kJ[/tex]
where,
The liquid is heated from 17.6 °C to 51.8 °C. We can calculate the heat required (Q₃) using the following expression.
[tex]Q_3 = c \times m \times \Delta T = \frac{1.75J}{g.\° C } \times 488.3g \times (51.8\° C-17.6\° C)) \times \frac{1kJ}{1000J} = 29.2 kJ[/tex]
The total amount of energy absorbed (Q) is the sum of the energy absorbed in each step.
[tex]Q = Q_1 + Q_2 + Q_3 = 58.8kJ+42.5kJ+29.2kJ= 130.5kJ[/tex]
In order to heat a 488.3 g solid, 58.8 kJ are required. To melt the solid, 42.5 kJ are required. To heat the liquid, 29.2 kJ are required. The total amount of energy absorbed is 130.5 kJ.
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