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The specific heat capacity is a key concept in solving calorimetry problems. It refers to the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius. In mathematical terms, it is represented by the formula: \[q = mc\Delta T\]

• q is the heat added in joules (J),
• m
is the mass in grams (g),
• \Delta T is the change in temperature in degrees Celsius (°C or K), and
• c
is the specific heat capacity in joules per gram per degree Celsius (J/g°C).

For example, water has a specific heat capacity of 4.18 J/g°C, while copper's is 0.385 J/g°C. This property allows us to calculate how much heat energy is needed to change the temperature of a substance.
In the problem, we need to calculate the required heat to raise both the copper pan and the water to their boiling points before summing the total heat required.

The boiling point is another critical term in this exercise. The boiling point of a substance is the temperature at which it changes from a liquid to a gas. For water, this is 100°C at standard atmospheric pressure.
To get water to its boiling point, you need to add enough thermal energy to overcome the heat of vaporization. This does not affect the pan, as we only need to account for the sensible heat required to bring it to 100°C.
In our exercise, we start with both the water and the copper pan at room temperature (25°C), so we must heat them up to the boiling point of water. This means calculating how much energy is needed for both components to raise their temperatures to 100°C without any phase change.

Heat transfer calculation involves determining how much heat energy is needed to change the state or temperature of a substance. The formula used here is: \[q = mc\Delta T\]
Let's break down the steps in solving the problem using this method:

• First, find the heat needed for the copper pan: \[q_{\text{pan}} = mc\Delta T\]\[= 300 \text{g} \times 0.385 J.g^{-1}.°C^{-1} \times (100 - 25)°C\]


• Second, calculate the heat needed for the water: \[q_{\text{water}} = mc\Delta T\] \[= 800 \text{g} \times 4.18 J.g^{-1}.°C^{-1} \times (100 - 25)°C\]

Finally, add both heat amounts to get the total heat energy:
\[q_{\text{total}} = q_{\text{pan}} + q_{\text{water}}\], where

• \[q_{\text{pan}}\] is the energy required to heat the copper pan
• \[q_{\text{water}}\] is for heating the water to boiling point.