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If 250g of Ni at 120 0C is dropped into 200g of water at 10 0C contained by a Calorimeter of 20 Cal / 0C heat Capacity, what will be the final temperature of the mixture (in 0C)? Given ( CN = .106 k Cal / 0c)
Correct answer is '22'. Can you explain this answer?
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If 250g of Ni at 120 0C is dropped into 200g of water at 10 0C contain...
Heat tost by Ni = heat gained by water + Calorimeter
Heat capacity of Ni, CN= 0.106 k cal /0c
So, 0.250 (0.106) (1200-1) = [ (10.200) (1.00 + 0.020) x (t - 100c)
3.18 -0.07t = 0.220 t-2.20
0.2471 = 5.38
t = 220c
Heat capacity of Ni, CN= 0.106 k cal /0c
So, 0.250 (0.106) (1200-1) = [ (10.200) (1.00 + 0.020) x (t - 100c)
3.18 -0.07t = 0.220 t-2.20
0.2471 = 5.38
t = 220c
Most Upvoted Answer
If 250g of Ni at 120 0C is dropped into 200g of water at 10 0C contain...
Heat tost by Ni = heat gained by water + Calorimeter
Heat capacity of Ni, CN= 0.106 k cal /0c
So, 0.250 (0.106) (1200-1) = [ (10.200) (1.00 + 0.020) x (t - 100c)
3.18 -0.07t = 0.220 t-2.20
0.2471 = 5.38
t = 220c
Heat capacity of Ni, CN= 0.106 k cal /0c
So, 0.250 (0.106) (1200-1) = [ (10.200) (1.00 + 0.020) x (t - 100c)
3.18 -0.07t = 0.220 t-2.20
0.2471 = 5.38
t = 220c
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Community Answer
If 250g of Ni at 120 0C is dropped into 200g of water at 10 0C contain...
Given:
- Mass of Ni (m1) = 250g
- Initial temperature of Ni (T1) = 120°C
- Mass of water (m2) = 200g
- Initial temperature of water (T2) = 10°C
- Heat capacity of calorimeter (C) = 20 cal/°C
- Heat capacity of water (Cw) = ?
- Heat capacity of Ni (Cn) = 0.106kcal/°C
- Final temperature of the mixture (Tf) = ?
Step 1: Calculate the heat lost by the nickel (Q1)
The heat lost by the nickel can be calculated using the formula:
Q1 = m1 * Cn * (T1 - Tf)
Step 2: Calculate the heat gained by the water (Q2)
The heat gained by the water can be calculated using the formula:
Q2 = m2 * Cw * (Tf - T2)
Step 3: Calculate the heat gained by the calorimeter (Q3)
The heat gained by the calorimeter can be calculated using the formula:
Q3 = C * (Tf - T2)
Step 4: Equate the heat lost and gained
Since the system is isolated, the heat lost by the nickel is equal to the heat gained by the water and calorimeter. Therefore, we can write:
Q1 = Q2 + Q3
Step 5: Substitute the values and solve for Tf
Substituting the values and rearranging the equation, we get:
m1 * Cn * (T1 - Tf) = m2 * Cw * (Tf - T2) + C * (Tf - T2)
Simplifying the equation further:
m1 * Cn * T1 - m1 * Cn * Tf = m2 * Cw * Tf - m2 * Cw * T2 + C * Tf - C * T2
Rearranging the terms and collecting like terms:
(m1 * Cn + m2 * Cw + C) * Tf = m1 * Cn * T1 + m2 * Cw * T2 + C * T2
Finally, solving for Tf:
Tf = (m1 * Cn * T1 + m2 * Cw * T2 + C * T2) / (m1 * Cn + m2 * Cw + C)
Step 6: Substitute the given values and calculate Tf
Substituting the given values into the equation:
Tf = (250g * 0.106kcal/°C * 120°C + 200g * Cw * 10°C + 20 cal/°C * 10°C) / (250g * 0.106kcal/°C + 200g * Cw + 20 cal/°C)
Since the heat capacity of water is not given, we cannot directly calculate Tf. However, we can assume that the heat capacity of water is much larger than the heat capacity of the calorimeter and nickel. Therefore, we can neglect the terms involving Cw and consider only the terms involving C and Cn.
Tf = (250g *
- Mass of Ni (m1) = 250g
- Initial temperature of Ni (T1) = 120°C
- Mass of water (m2) = 200g
- Initial temperature of water (T2) = 10°C
- Heat capacity of calorimeter (C) = 20 cal/°C
- Heat capacity of water (Cw) = ?
- Heat capacity of Ni (Cn) = 0.106kcal/°C
- Final temperature of the mixture (Tf) = ?
Step 1: Calculate the heat lost by the nickel (Q1)
The heat lost by the nickel can be calculated using the formula:
Q1 = m1 * Cn * (T1 - Tf)
Step 2: Calculate the heat gained by the water (Q2)
The heat gained by the water can be calculated using the formula:
Q2 = m2 * Cw * (Tf - T2)
Step 3: Calculate the heat gained by the calorimeter (Q3)
The heat gained by the calorimeter can be calculated using the formula:
Q3 = C * (Tf - T2)
Step 4: Equate the heat lost and gained
Since the system is isolated, the heat lost by the nickel is equal to the heat gained by the water and calorimeter. Therefore, we can write:
Q1 = Q2 + Q3
Step 5: Substitute the values and solve for Tf
Substituting the values and rearranging the equation, we get:
m1 * Cn * (T1 - Tf) = m2 * Cw * (Tf - T2) + C * (Tf - T2)
Simplifying the equation further:
m1 * Cn * T1 - m1 * Cn * Tf = m2 * Cw * Tf - m2 * Cw * T2 + C * Tf - C * T2
Rearranging the terms and collecting like terms:
(m1 * Cn + m2 * Cw + C) * Tf = m1 * Cn * T1 + m2 * Cw * T2 + C * T2
Finally, solving for Tf:
Tf = (m1 * Cn * T1 + m2 * Cw * T2 + C * T2) / (m1 * Cn + m2 * Cw + C)
Step 6: Substitute the given values and calculate Tf
Substituting the given values into the equation:
Tf = (250g * 0.106kcal/°C * 120°C + 200g * Cw * 10°C + 20 cal/°C * 10°C) / (250g * 0.106kcal/°C + 200g * Cw + 20 cal/°C)
Since the heat capacity of water is not given, we cannot directly calculate Tf. However, we can assume that the heat capacity of water is much larger than the heat capacity of the calorimeter and nickel. Therefore, we can neglect the terms involving Cw and consider only the terms involving C and Cn.
Tf = (250g *
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If 250g of Ni at 120 0C is dropped into 200g of water at 10 0C contained by a Calorimeter of 20 Cal / 0C heat Capacity, what will be the final temperature of the mixture (in 0C)? Given ( CN = .106 k Cal / 0c)Correct answer is '22'. Can you explain this answer?
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If 250g of Ni at 120 0C is dropped into 200g of water at 10 0C contained by a Calorimeter of 20 Cal / 0C heat Capacity, what will be the final temperature of the mixture (in 0C)? Given ( CN = .106 k Cal / 0c)Correct answer is '22'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about If 250g of Ni at 120 0C is dropped into 200g of water at 10 0C contained by a Calorimeter of 20 Cal / 0C heat Capacity, what will be the final temperature of the mixture (in 0C)? Given ( CN = .106 k Cal / 0c)Correct answer is '22'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If 250g of Ni at 120 0C is dropped into 200g of water at 10 0C contained by a Calorimeter of 20 Cal / 0C heat Capacity, what will be the final temperature of the mixture (in 0C)? Given ( CN = .106 k Cal / 0c)Correct answer is '22'. Can you explain this answer?.
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If 250g of Ni at 120 0C is dropped into 200g of water at 10 0C contained by a Calorimeter of 20 Cal / 0C heat Capacity, what will be the final temperature of the mixture (in 0C)? Given ( CN = .106 k Cal / 0c)Correct answer is '22'. Can you explain this answer?, a detailed solution for If 250g of Ni at 120 0C is dropped into 200g of water at 10 0C contained by a Calorimeter of 20 Cal / 0C heat Capacity, what will be the final temperature of the mixture (in 0C)? Given ( CN = .106 k Cal / 0c)Correct answer is '22'. Can you explain this answer? has been provided alongside types of If 250g of Ni at 120 0C is dropped into 200g of water at 10 0C contained by a Calorimeter of 20 Cal / 0C heat Capacity, what will be the final temperature of the mixture (in 0C)? Given ( CN = .106 k Cal / 0c)Correct answer is '22'. Can you explain this answer? theory, EduRev gives you an
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