Understanding Heat Transfer in a Copper Block

This question asks us to calculate the amount of heat transferred to a block of copper when its temperature increases. This involves the concept of specific heat capacity, which is the amount of heat required to raise the temperature of one unit of mass of a substance by one degree Celsius (or Kelvin).

Key Concepts for Calculating Heat Transfer

• Heat Transfer (\(Q\)): The energy transferred from one object to another due to a temperature difference.
• Mass (\(m\)): The amount of matter in the object, typically measured in kilograms (kg).
• Specific Heat Capacity (\(c\)): A physical property of a substance that quantifies the amount of heat energy needed to raise the temperature of 1 kg of the substance by 1 K (or 1 °C). It is usually measured in Jkg^-1K^-1 or Jkg^-1°C^-1.
• Temperature Change (\(\Delta T\)): The difference between the final and initial temperatures. It is the same value whether measured in °C or K.

The formula used to calculate the heat transferred (\(Q\)) to a substance when its temperature changes is:

\( Q = mc\Delta T \)

Where:

• \( Q \) is the heat transferred (in Joules, J)
• \( m \) is the mass of the substance (in kilograms, kg)
• \( c \) is the specific heat capacity of the substance (in Jkg^-1K^-1)
• \( \Delta T \) is the change in temperature (in Kelvin, K, or °C)

Step-by-Step Calculation of Heat Transferred

Let's identify the given values in the problem:

• Mass of the copper block, \( m = 50 \, \text{g} \). We need to convert this to kilograms: \( m = 50 \, \text{g} \times \frac{1 \, \text{kg}}{1000 \, \text{g}} = 0.050 \, \text{kg} \).
• Initial temperature, \( T_i = 20^\circ \text{C} \).
• Final temperature, \( T_f = 60^\circ \text{C} \).
• Specific heat capacity of copper, \( c = 386 \, \text{Jkg}^{-1}\text{K}^{-1} \).

Now, calculate the change in temperature, \( \Delta T \):

\( \Delta T = T_f - T_i = 60^\circ \text{C} - 20^\circ \text{C} = 40^\circ \text{C} \)

Since a change in Celsius is equivalent to a change in Kelvin, \( \Delta T = 40 \, \text{K} \).

Finally, use the formula \( Q = mc\Delta T \) to calculate the heat transferred:

\( Q = (0.050 \, \text{kg})(386 \, \text{Jkg}^{-1}\text{K}^{-1})(40 \, \text{K}) \)

\( Q = 0.050 \times 386 \times 40 \, \text{J} \)

\( Q = 2.0 \times 386 \, \text{J} \)

\( Q = 772 \, \text{J} \)

So, the amount of heat transferred to the copper block is 772 Joules.

Final Result Confirmation

The calculated heat transferred is 772 J, which matches one of the given options.

Revision Table: Heat Transfer Formula Review

Additional Information: Specific Heat Capacity and Materials

Specific heat capacity is an important property that varies greatly among different substances. It tells us how much energy is needed to change the temperature of a substance. Materials with high specific heat capacity, like water, require a lot of energy to change their temperature. Materials with low specific heat capacity, like metals (including copper), heat up or cool down more quickly when the same amount of heat is added or removed.

This property is why water is used in cooling systems (it absorbs a lot of heat without a large temperature rise) and why metal pots heat up quickly on a stove.

The units of specific heat capacity, Jkg^-1K^-1 or Jkg^-1°C^-1, are equivalent because a temperature difference of 1 K is equal to a temperature difference of 1 °C.