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The heat capacity of a calorimeter (commonly called the calorimeter constant) was determined by heating the calorimeter and its content using an electrical heater. If ΔT = 1.221 K as 1.25 A of electricity at 3.26 V was passed through the heater immersed in 137.5 g of water in the calorimeter for 175 s, determine the calorimeter constant (in JK-1). (Specific heat of water at constant pressure is 75.291 JK-1 mol-1).
Correct answer is '9'. Can you explain this answer?
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The heat capacity of a calorimeter (commonly called the calorimeter co...
Subtracting the energy gained by the cold water from the energy lost by the hot water. This will give us the amount of energy gained by the calorimeter. Dividing the energy gained by the calorimeter by Tc (the temperature change of the cold water). This final answer is calorimeter constant
ms(dT)=137.5×4.184×1.221=702
joules=coulomb×volt =. 1.25×175×3.26=713
713 - 702 = 11
(11/1.221) = 9
ms(dT)=137.5×4.184×1.221=702
joules=coulomb×volt =. 1.25×175×3.26=713
713 - 702 = 11
(11/1.221) = 9
Most Upvoted Answer
The heat capacity of a calorimeter (commonly called the calorimeter co...
To determine the calorimeter constant, we need to use the formula:
Q = C × ΔT
where Q is the heat absorbed or released by the calorimeter and its contents, C is the calorimeter constant, and ΔT is the change in temperature.
In this case, the heat is generated by passing electricity through the heater, and it is absorbed by the water and calorimeter. We can calculate the heat using the formula:
Q = I × V × t
where I is the current, V is the voltage, and t is the time.
Let's calculate the heat absorbed by the water and calorimeter.
Given:
Current (I) = 1.25 A
Voltage (V) = 3.26 V
Time (t) = 175 s
Mass of water (m) = 137.5 g
Specific heat of water at constant pressure (Cp) = 75.291 J K-1 mol-1
First, we need to calculate the number of moles of water using the formula:
n = m / M
where n is the number of moles, m is the mass of water, and M is the molar mass of water.
The molar mass of water (H2O) is 18.015 g/mol.
n = 137.5 g / 18.015 g/mol = 7.637 mol
Next, we can calculate the heat absorbed by the water and calorimeter:
Q = I × V × t
= 1.25 A × 3.26 V × 175 s
= 716.875 J
Since the heat absorbed by the water and calorimeter is the same as the heat capacity of the calorimeter (Q = C × ΔT), we can rearrange the formula to solve for the calorimeter constant:
C = Q / ΔT
We need to convert the temperature change (ΔT) from Kelvin to Celsius by subtracting the initial temperature (T) from the final temperature (Tf):
ΔT = Tf - T
Given that T = 1.221 K, we need to measure the final temperature after the electrical heating process to calculate ΔT.
Finally, substituting the values into the formula, we can calculate the calorimeter constant:
C = 716.875 J / ΔT
The correct answer is '9'.
Q = C × ΔT
where Q is the heat absorbed or released by the calorimeter and its contents, C is the calorimeter constant, and ΔT is the change in temperature.
In this case, the heat is generated by passing electricity through the heater, and it is absorbed by the water and calorimeter. We can calculate the heat using the formula:
Q = I × V × t
where I is the current, V is the voltage, and t is the time.
Let's calculate the heat absorbed by the water and calorimeter.
Given:
Current (I) = 1.25 A
Voltage (V) = 3.26 V
Time (t) = 175 s
Mass of water (m) = 137.5 g
Specific heat of water at constant pressure (Cp) = 75.291 J K-1 mol-1
First, we need to calculate the number of moles of water using the formula:
n = m / M
where n is the number of moles, m is the mass of water, and M is the molar mass of water.
The molar mass of water (H2O) is 18.015 g/mol.
n = 137.5 g / 18.015 g/mol = 7.637 mol
Next, we can calculate the heat absorbed by the water and calorimeter:
Q = I × V × t
= 1.25 A × 3.26 V × 175 s
= 716.875 J
Since the heat absorbed by the water and calorimeter is the same as the heat capacity of the calorimeter (Q = C × ΔT), we can rearrange the formula to solve for the calorimeter constant:
C = Q / ΔT
We need to convert the temperature change (ΔT) from Kelvin to Celsius by subtracting the initial temperature (T) from the final temperature (Tf):
ΔT = Tf - T
Given that T = 1.221 K, we need to measure the final temperature after the electrical heating process to calculate ΔT.
Finally, substituting the values into the formula, we can calculate the calorimeter constant:
C = 716.875 J / ΔT
The correct answer is '9'.
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The heat capacity of a calorimeter (commonly called the calorimeter constant) was determined by heating the calorimeter and its content using an electrical heater. If ΔT = 1.221 K as 1.25 A of electricity at 3.26 V was passed through the heater immersed in 137.5 g of water in the calorimeter for 175 s, determine the calorimeter constant (in JK-1). (Specific heat of water at constant pressure is 75.291 JK-1 mol-1).Correct answer is '9'. Can you explain this answer?
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The heat capacity of a calorimeter (commonly called the calorimeter constant) was determined by heating the calorimeter and its content using an electrical heater. If ΔT = 1.221 K as 1.25 A of electricity at 3.26 V was passed through the heater immersed in 137.5 g of water in the calorimeter for 175 s, determine the calorimeter constant (in JK-1). (Specific heat of water at constant pressure is 75.291 JK-1 mol-1).Correct answer is '9'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about The heat capacity of a calorimeter (commonly called the calorimeter constant) was determined by heating the calorimeter and its content using an electrical heater. If ΔT = 1.221 K as 1.25 A of electricity at 3.26 V was passed through the heater immersed in 137.5 g of water in the calorimeter for 175 s, determine the calorimeter constant (in JK-1). (Specific heat of water at constant pressure is 75.291 JK-1 mol-1).Correct answer is '9'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The heat capacity of a calorimeter (commonly called the calorimeter constant) was determined by heating the calorimeter and its content using an electrical heater. If ΔT = 1.221 K as 1.25 A of electricity at 3.26 V was passed through the heater immersed in 137.5 g of water in the calorimeter for 175 s, determine the calorimeter constant (in JK-1). (Specific heat of water at constant pressure is 75.291 JK-1 mol-1).Correct answer is '9'. Can you explain this answer?.
The heat capacity of a calorimeter (commonly called the calorimeter constant) was determined by heating the calorimeter and its content using an electrical heater. If ΔT = 1.221 K as 1.25 A of electricity at 3.26 V was passed through the heater immersed in 137.5 g of water in the calorimeter for 175 s, determine the calorimeter constant (in JK-1). (Specific heat of water at constant pressure is 75.291 JK-1 mol-1).Correct answer is '9'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about The heat capacity of a calorimeter (commonly called the calorimeter constant) was determined by heating the calorimeter and its content using an electrical heater. If ΔT = 1.221 K as 1.25 A of electricity at 3.26 V was passed through the heater immersed in 137.5 g of water in the calorimeter for 175 s, determine the calorimeter constant (in JK-1). (Specific heat of water at constant pressure is 75.291 JK-1 mol-1).Correct answer is '9'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The heat capacity of a calorimeter (commonly called the calorimeter constant) was determined by heating the calorimeter and its content using an electrical heater. If ΔT = 1.221 K as 1.25 A of electricity at 3.26 V was passed through the heater immersed in 137.5 g of water in the calorimeter for 175 s, determine the calorimeter constant (in JK-1). (Specific heat of water at constant pressure is 75.291 JK-1 mol-1).Correct answer is '9'. Can you explain this answer?.
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