A 155 -g sample of copper was heated to \(150.0^{\circ} \mathrm{C}\), then placed into \(250.0 \mathrm{~g}\) water at \(19.8^{\circ} \mathrm{C}\). (The specific heat of copper is \(0.385\) \(\mathrm{J} / \mathrm{g}^{\circ} \mathrm{C}\).) Calculate the final temperature of the mixture. (Assume no heat is lost to the surroundings.)

Understanding specific heat capacity is crucial when solving heat transfer problems. Specific heat capacity, often simply called specific heat, is the amount of heat energy required to raise the temperature of one gram of a substance by one degree Celsius (or one Kelvin).

It is a characteristic property of each material and is essential for determining how much heat energy different substances can absorb or release. For instance, in the provided exercise, the specific heat of copper is given as 0.385 J/g°C. This means that for every gram of copper, 0.385 joules of heat is needed to increase its temperature by one degree Celsius.

Different substances have different specific heats, which explains why some materials heat up or cool down more quickly than others. Water, for example, has a relatively high specific heat of 4.184 J/g°C, which is why it takes a longer time to heat up and also retains heat longer once it is warm.

Thermal equilibrium is a key concept in thermodynamics and plays a vital role in heat transfer problems. It refers to the condition where two substances in contact with each other exchange heat until they reach the same temperature.

At this point, there is no further heat flow between the substances, as the rates of heat transfer equalize. In the context of the exercise, when the hot copper sample is placed into the cooler water, heat flows from the copper to the water until both materials reach the same final temperature. This final temperature is what the problem asks us to calculate.

Thermal equilibrium is a fundamental principle that underpins the idea of heat transfer—without it, we could not make accurate predictions or calculations regarding how substances will interact when they exchange heat. Understanding this concept helps students to comprehend the principle that 'heat lost = heat gained' in an isolated system, which is central to addressing these types of exercises.

The heat exchange formula is an invaluable tool for solving heat transfer problems and it's grounded in the principle of conservation of energy. The basic formula is given by: \( q = mc\Delta T \) where \( q \) represents the heat transferred, \( m \) is the mass of the substance, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature.

When a hot substance cools down, it loses heat, and a cooler substance absorbs that heat when it warms up. In the case of the exercise involving copper and water, by setting the heat loss of the copper equal to the heat gain of the water, we can calculate the final temperature of both substances when they reach thermal equilibrium. This involves setting up an equation that expresses this balance and solving for the unknown final temperature.

Improving understanding of this concept enables students to handle problems involving multiple substances and varying conditions, as it provides a reliable method for quantifying the energy exchange during heating or cooling processes.