An unknown material with a mass of 1 kg is heated from 200C to 500C. The amount of heat energy required was 3.7 kJ. Calculate the specific heat and identify the material.
An unknown material with a mass of 1 kg is heated from 200C to 500C. The amount of heat energy required was 3.7 kJ. Calculate the specific heat and identify the material.
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![**Transcription:**
An unknown material with a mass of 1 kg is heated from 20°C to 50°C. The amount of heat energy required was 3.7 kJ. Calculate the specific heat and identify the material.
**Explanation for Educational Context:**
This problem involves calculating the specific heat capacity of a material using the given quantities. The specific heat capacity is a property that indicates the amount of heat required to change the temperature of a unit mass of a substance by one degree Celsius.
Given:
- Mass (m) = 1 kg
- Initial Temperature (T₁) = 20°C
- Final Temperature (T₂) = 50°C
- Heat Energy (Q) = 3.7 kJ
You can find the specific heat capacity (c) using the formula:
\[ Q = m \cdot c \cdot \Delta T \]
Where:
- \( \Delta T = T₂ - T₁ \) is the change in temperature.
Substituting the given values into the formula allows for solving the specific heat and ultimately identifying the material based on known specific heat values for various substances.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fadbb6fa3-6b43-4484-8eb2-c92bfed12f92%2F8b9f1d75-1c40-4af5-98f3-19513f0e40b6%2F3vh94r8_processed.png&w=3840&q=75)
Transcribed Image Text:**Transcription:**
An unknown material with a mass of 1 kg is heated from 20°C to 50°C. The amount of heat energy required was 3.7 kJ. Calculate the specific heat and identify the material.
**Explanation for Educational Context:**
This problem involves calculating the specific heat capacity of a material using the given quantities. The specific heat capacity is a property that indicates the amount of heat required to change the temperature of a unit mass of a substance by one degree Celsius.
Given:
- Mass (m) = 1 kg
- Initial Temperature (T₁) = 20°C
- Final Temperature (T₂) = 50°C
- Heat Energy (Q) = 3.7 kJ
You can find the specific heat capacity (c) using the formula:
\[ Q = m \cdot c \cdot \Delta T \]
Where:
- \( \Delta T = T₂ - T₁ \) is the change in temperature.
Substituting the given values into the formula allows for solving the specific heat and ultimately identifying the material based on known specific heat values for various substances.
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