22) A 4.0-kg aluminum block is originally at 10°C. If 160 kJ of heat is added to the block, what is its final temperature? The specific heat capacity of aluminum is 910J/kg. K. A) 34°C B) 54°C C) 44°C D) 24°C
Energy transfer
The flow of energy from one region to another region is referred to as energy transfer. Since energy is quantitative; it must be transferred to a body or a material to work or to heat the system.
Molar Specific Heat
Heat capacity is the amount of heat energy absorbed or released by a chemical substance per the change in temperature of that substance. The change in heat is also called enthalpy. The SI unit of heat capacity is Joules per Kelvin, which is (J K-1)
Thermal Properties of Matter
Thermal energy is described as one of the form of heat energy which flows from one body of higher temperature to the other with the lower temperature when these two bodies are placed in contact to each other. Heat is described as the form of energy which is transferred between the two systems or in between the systems and their surrounding by the virtue of difference in temperature. Calorimetry is that branch of science which helps in measuring the changes which are taking place in the heat energy of a given body.
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![### Physics Problem on Heat Transfer
**Question 22:**
A 4.0-kg aluminum block is originally at 10°C. If 160 kJ of heat is added to the block, what is its final temperature? The specific heat capacity of aluminum is 910 J/kg·K.
**Answer Choices:**
A) 34°C
B) 54°C
C) 44°C
D) 24°C
**Detailed Explanation:**
To solve this problem, we can use the formula for heat transfer:
\[ Q = mc\Delta T \]
Where:
- \( Q \) is the heat added (in joules, J)
- \( m \) is the mass of the substance (in kilograms, kg)
- \( c \) is the specific heat capacity of the substance (in joules per kilogram per kelvin, J/kg·K)
- \( \Delta T \) is the change in temperature (in kelvin or degrees Celsius, K or °C)
First, convert the heat added from kJ to J:
\[ Q = 160 \, \text{kJ} = 160,000 \, \text{J} \]
Given:
\[ m = 4.0 \, \text{kg} \]
\[ c = 910 \, \text{J/kg·K} \]
Rearrange the formula to solve for the change in temperature (\(\Delta T\)):
\[ \Delta T = \frac{Q}{mc} \]
Calculate \(\Delta T\):
\[ \Delta T = \frac{160,000 \, \text{J}}{(4.0 \, \text{kg}) (910 \, \text{J/kg·K})} \]
\[ \Delta T = \frac{160,000}{3,640} \]
\[ \Delta T = 44 \, \text{°C} \]
Finally, add the change in temperature to the initial temperature to find the final temperature:
\[ T_{\text{final}} = T_{\text{initial}} + \Delta T \]
\[ T_{\text{final}} = 10 \, \text{°C} + 44 \, \text{°C} \]
\[ T_{\text{final}} = 54 \, \text{°C} \]
Therefore, the correct answer is:
B) 54°C](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6527be19-4214-4a2f-aca2-4f82253fbdd1%2F221a40a0-a1a0-48bd-894c-ae5507d7e99a%2F479hiet_processed.png&w=3840&q=75)
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