Calculating Heat Capacity of a Steel Vessel

The question asks us to find the heat capacity of a steel vessel given its mass, the specific heat capacity of steel, and a temperature change. While the temperature rise is given (10 degrees), it is actually not needed to calculate the heat capacity itself. The heat capacity is a property of the object (the steel vessel) and depends on its mass and the material's specific heat capacity.

Understanding Heat Capacity and Specific Heat Capacity

Let's first understand the key terms:

• Specific Heat Capacity (\(c\)): This is an intrinsic property of a substance. It is defined as the amount of heat energy required to raise the temperature of 1 kilogram of that substance by 1 degree Celsius or 1 Kelvin. Its standard unit is Joules per kilogram per Kelvin (\(Jkg^{-1}K^{-1}\)) or Joules per kilogram per degree Celsius (\(Jkg^{-1}C^{-1}\)). For steel, it's given as \(500 Jkg^{-1}K^{-1}\).
• Heat Capacity (\(C\)): This is a property of a specific object. It is defined as the amount of heat energy required to raise the temperature of the entire object by 1 degree Celsius or 1 Kelvin. Its standard unit is Joules per Kelvin (\(JK^{-1}\)) or Joules per degree Celsius (\(JC^{-1}\)).

The heat capacity (\(C\)) of an object is related to its mass (\(m\)) and the specific heat capacity (\(c\)) of the material it is made from by the formula:

\[C = m \times c\]

This formula tells us that a more massive object made of the same material will have a higher heat capacity, meaning it requires more energy to change its temperature by the same amount.

Applying the Formula to the Steel Vessel

We are given the following information for the steel vessel:

• Mass of the steel vessel, \(m = 2.5\) kg
• Specific heat capacity of steel, \(c = 500\) \(Jkg^{-1}K^{-1}\)

Using the formula \(C = m \times c\), we can calculate the heat capacity of the steel vessel:

\[C = (2.5 \text{ kg}) \times (500 \text{ Jkg}^{-1}K^{-1})\]

Let's perform the multiplication:

\[C = 2.5 \times 500 \text{ JK}^{-1}\]

\[C = 1250 \text{ JK}^{-1}\]

So, the heat capacity of the steel vessel is \(1250 \text{ JK}^{-1}\).

Analyzing the Options

Let's compare our calculated value with the given options:

• Option 1: \(200 Jkg^{-1}K^{-1}\) - This unit (\(Jkg^{-1}K^{-1}\)) is for specific heat capacity, not heat capacity (\(JK^{-1}\)). Also, the value is incorrect.
• Option 2: \(1,250 JK^{-1}\) - This matches our calculated value and has the correct unit for heat capacity.
• Option 3: \(125 JK^{-1}\) - This value is incorrect. It seems like a calculation error might have occurred (e.g., 2.5 * 50 = 125).
• Option 4: \(20 Jkg^{-1}K^{-1}\) - This unit (\(Jkg^{-1}K^{-1}\)) is for specific heat capacity, not heat capacity (\(JK^{-1}\)). Also, the value is incorrect.

Based on our calculation, the correct heat capacity of the steel vessel is \(1250 JK^{-1}\).

Conclusion

The heat capacity of the steel vessel of mass 2.5 kg with a specific heat capacity of 500 \(Jkg^{-1}K^{-1}\) is \(1250 JK^{-1}\).

Revision Table: Heat Capacity Calculations

Additional Information: Factors Affecting Heat Capacity

The heat capacity of an object is an extensive property, meaning it depends on the amount of substance present. Here are some factors that influence the heat capacity of an object:

• Mass (\(m\)): As shown in the formula \(C = m \times c\), heat capacity is directly proportional to mass. A larger mass requires more heat to change its temperature by the same amount.
• Material (Specific Heat Capacity, \(c\)): Different materials have different specific heat capacities. For instance, water has a very high specific heat capacity compared to metals like steel. This means water can absorb a lot more heat energy than an equal mass of steel for the same temperature rise.
• Phase of the Material: The specific heat capacity of a substance can change depending on whether it is in a solid, liquid, or gaseous state. For example, the specific heat capacity of water is different from that of ice or steam.

Understanding heat capacity is crucial in various applications, from designing heating and cooling systems to studying climate patterns, where the high heat capacity of oceans plays a significant role in stabilizing Earth's temperature.