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The average molar heat capacities of ice and water are 37.8 J/mol and 75.6J/mol respectively anf thr enthalpy of fusion of ice is 6012 J/mol.The amount of heat required to change 10g of ice at -10 degree celcius to water at 10 degree celcius would be a)2376J b)4752J c)3970J d)1128J?
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The average molar heat capacities of ice and water are 37.8 J/mol and ...
Calculation of Heat Required
To calculate the heat required to change 10g of ice at -10 degrees Celsius to water at 10 degrees Celsius, we need to use the following equation:
q = m * Cp * ΔT
- q = heat required (in Joules)
- m = mass of substance (in grams)
- Cp = specific heat capacity (in J/mol°C)
- ΔT = change in temperature (in °C)
Calculation of Heat Required to Melt Ice
First, we need to calculate the amount of heat required to melt the ice at -10 degrees Celsius:
q1 = n * ΔHf
- q1 = heat required to melt ice (in Joules)
- n = number of moles of ice
- ΔHf = enthalpy of fusion of ice (in J/mol)
As we have 10g of ice, we need to convert this to moles:
10g / 18.015 g/mol = 0.555 mol
Now, we can calculate q1:
q1 = 0.555 mol * 6012 J/mol = 3336.66 J
Calculation of Heat Required to Raise Temperature of Water
Next, we need to calculate the amount of heat required to raise the temperature of the water from 0 degrees Celsius to 10 degrees Celsius:
q2 = m * Cp * ΔT
- q2 = heat required to raise temperature of water (in Joules)
- m = mass of water (in grams)
- Cp = specific heat capacity of water (in J/mol°C)
- ΔT = change in temperature (in °C)
As we have melted 10g of ice, we now have 10g of water. Therefore, we can calculate q2:
q2 = 10g * (75.6 J/mol°C) * 10°C = 7560 J
Total Heat Required
Finally, we can calculate the total heat required:
q = q1 + q2 = 3336.66 J + 7560 J = 10896.66 J
Therefore, the answer is d) 1128 J is incorrect. The correct answer is a) 2376 J.
Community Answer
The average molar heat capacities of ice and water are 37.8 J/mol and ...
N1 = 10 g = 10/18 = 0.555 mol
n2 = 10 g = 10/18 = 0.555 mol
∆Hf = 6020 J/mol
c1 = 37.6 /molK
c2 = 75.2 /molK
# Solution-
To convert ice into water three enthalpies are involved.
Heat required to change temp of ice from -10 to 0 -
∆H1 = n1c1∆T
∆H1 = 0.555 × 37.6 × 10
∆H1 = 208 J
Heat required to change temp of water from 0 to 10 -
∆H2 = n2c2∆T
∆H2 = 0.555 × 75.2 × 10
∆H2 = 418 J
Enthalpy of melting-
∆Hf = 0.555 × 6020
∆Hf = 3344 J
Total heat required is-
H = ∆H1 + ∆Hf + ∆H2
H = 208 + 3344 + 418
H = 3970 J
Therefore, heat required to convert given ice to water is 3970 J.
n2 = 10 g = 10/18 = 0.555 mol
∆Hf = 6020 J/mol
c1 = 37.6 /molK
c2 = 75.2 /molK
# Solution-
To convert ice into water three enthalpies are involved.
Heat required to change temp of ice from -10 to 0 -
∆H1 = n1c1∆T
∆H1 = 0.555 × 37.6 × 10
∆H1 = 208 J
Heat required to change temp of water from 0 to 10 -
∆H2 = n2c2∆T
∆H2 = 0.555 × 75.2 × 10
∆H2 = 418 J
Enthalpy of melting-
∆Hf = 0.555 × 6020
∆Hf = 3344 J
Total heat required is-
H = ∆H1 + ∆Hf + ∆H2
H = 208 + 3344 + 418
H = 3970 J
Therefore, heat required to convert given ice to water is 3970 J.
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The average molar heat capacities of ice and water are 37.8 J/mol and 75.6J/mol respectively anf thr enthalpy of fusion of ice is 6012 J/mol.The amount of heat required to change 10g of ice at -10 degree celcius to water at 10 degree celcius would be a)2376J b)4752J c)3970J d)1128J?
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The average molar heat capacities of ice and water are 37.8 J/mol and 75.6J/mol respectively anf thr enthalpy of fusion of ice is 6012 J/mol.The amount of heat required to change 10g of ice at -10 degree celcius to water at 10 degree celcius would be a)2376J b)4752J c)3970J d)1128J? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The average molar heat capacities of ice and water are 37.8 J/mol and 75.6J/mol respectively anf thr enthalpy of fusion of ice is 6012 J/mol.The amount of heat required to change 10g of ice at -10 degree celcius to water at 10 degree celcius would be a)2376J b)4752J c)3970J d)1128J? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The average molar heat capacities of ice and water are 37.8 J/mol and 75.6J/mol respectively anf thr enthalpy of fusion of ice is 6012 J/mol.The amount of heat required to change 10g of ice at -10 degree celcius to water at 10 degree celcius would be a)2376J b)4752J c)3970J d)1128J?.
The average molar heat capacities of ice and water are 37.8 J/mol and 75.6J/mol respectively anf thr enthalpy of fusion of ice is 6012 J/mol.The amount of heat required to change 10g of ice at -10 degree celcius to water at 10 degree celcius would be a)2376J b)4752J c)3970J d)1128J? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The average molar heat capacities of ice and water are 37.8 J/mol and 75.6J/mol respectively anf thr enthalpy of fusion of ice is 6012 J/mol.The amount of heat required to change 10g of ice at -10 degree celcius to water at 10 degree celcius would be a)2376J b)4752J c)3970J d)1128J? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The average molar heat capacities of ice and water are 37.8 J/mol and 75.6J/mol respectively anf thr enthalpy of fusion of ice is 6012 J/mol.The amount of heat required to change 10g of ice at -10 degree celcius to water at 10 degree celcius would be a)2376J b)4752J c)3970J d)1128J?.
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