21) How much heat is required to raise the temperature of a 225-g lead ball from 15.0°C to 25.0°C? The specific heat of lead is 128 J/kg.K. A) 725 J B) 576 J C) 217J D) 288 J E) 145 J
21) How much heat is required to raise the temperature of a 225-g lead ball from 15.0°C to 25.0°C? The specific heat of lead is 128 J/kg.K. A) 725 J B) 576 J C) 217J D) 288 J E) 145 J
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter16: Temperature And The Kinetic Theory Of Gases
Section: Chapter Questions
Problem 3P
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Please show all calculations and type them out since I have bad eyesight and it is very difficult for me to read handwriting
![### Heat Required to Raise the Temperature of Lead
**Problem Statement:**
How much heat is required to raise the temperature of a 225-g lead ball from 15.0°C to 25.0°C? The specific heat of lead is 128 J/(kg·K).
**Multiple Choice Answers:**
A) 725 J
B) 576 J
C) 217 J
D) 288 J
E) 145 J
**Step-By-Step Solution:**
To solve this problem, we use the formula for heat transfer:
\[ Q = mcΔT \]
Where:
- \( Q \) is the heat required (in joules, J)
- \( m \) is the mass (in kilograms, kg)
- \( c \) is the specific heat capacity (in joules per kilogram per kelvin, J/(kg·K))
- \( ΔT \) is the change in temperature (in degrees Celsius or Kelvin, as \( 1°C = 1 K \))
1. **Convert the mass from grams to kilograms:**
\[ 225\,g = 0.225\,kg \]
2. **Specific heat capacity of lead:**
\[ c = 128\,J/(kg·K) \]
3. **Change in temperature:**
\[ ΔT = 25.0°C - 15.0°C = 10.0\,K \]
4. **Calculate the heat required:**
\[ Q = (0.225\,kg) \times (128\,J/(kg·K)) \times (10.0\,K) \]
\[ Q = 288\,J \]
Thus, the heat required to raise the temperature of the lead ball is 288 J.
**Correct answer:**
D) 288 J](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6527be19-4214-4a2f-aca2-4f82253fbdd1%2Fb2addb8d-8284-43b5-9448-bd320bc6cd5a%2Fooz27p_processed.png&w=3840&q=75)
Transcribed Image Text:### Heat Required to Raise the Temperature of Lead
**Problem Statement:**
How much heat is required to raise the temperature of a 225-g lead ball from 15.0°C to 25.0°C? The specific heat of lead is 128 J/(kg·K).
**Multiple Choice Answers:**
A) 725 J
B) 576 J
C) 217 J
D) 288 J
E) 145 J
**Step-By-Step Solution:**
To solve this problem, we use the formula for heat transfer:
\[ Q = mcΔT \]
Where:
- \( Q \) is the heat required (in joules, J)
- \( m \) is the mass (in kilograms, kg)
- \( c \) is the specific heat capacity (in joules per kilogram per kelvin, J/(kg·K))
- \( ΔT \) is the change in temperature (in degrees Celsius or Kelvin, as \( 1°C = 1 K \))
1. **Convert the mass from grams to kilograms:**
\[ 225\,g = 0.225\,kg \]
2. **Specific heat capacity of lead:**
\[ c = 128\,J/(kg·K) \]
3. **Change in temperature:**
\[ ΔT = 25.0°C - 15.0°C = 10.0\,K \]
4. **Calculate the heat required:**
\[ Q = (0.225\,kg) \times (128\,J/(kg·K)) \times (10.0\,K) \]
\[ Q = 288\,J \]
Thus, the heat required to raise the temperature of the lead ball is 288 J.
**Correct answer:**
D) 288 J
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