Problem 2 - General Chemistry: Principles and Modern Applications

Question

Calculate the final temperature that results when (a) a 12.6 g sample of water at \(22.9^{\circ} \mathrm{C}\) absorbs \(875 \mathrm{J}\) of heat; (b) a 1.59 kg sample of platinum at \(78.2^{\circ} \mathrm{C}\) gives off \(1.05 \mathrm{kcal}\) of heat \(\left(\mathrm{sp} \mathrm{ht} \text { of } \mathrm{Pt}=0.032 \mathrm{cal} \mathrm{g}^{-1}\right.\) \(\left.^{\circ} \mathrm{C}^{-1}\right)\).

Step-by-Step Solution

Answer

The final temperature of the water after it absorbs 875 J of heat is calculated to be approximately 37.8°C. The final temperature of the platinum after it gives off 1.05 kcal of heat is calculated to be approximately 76.3°C.

Step 1: Calculating Final Temperature of Heated Water

To calculate the final temperature after heating the water, the heat capacity equation is used. It is given that the water absorbs 875 J of heat, the mass is 12.6 g, the initial temperature is 22.9°C, and the specific heat of water is 4.18 J/g°C. The equation will be set up as \(875 = 12.6 \times 4.18 \times (T_f - 22.9)\), where \(T_f\) is the final temperature. Solving for \(T_f\) will give the final temperature of the water after heating.

Step 2: Calculating Final Temperature of Cooled Platinum

To calculate the final temperature after cooling the platinum, the heat capacity equation is used again. It is given that the platinum gives off 1.05 kcal of heat (which needs to be converted to joules for consistent units), the mass is 1.59 kg (which needs to be converted to grams for consistent units), the initial temperature is 78.2°C, and the specific heat of platinum is 0.032 cal/g°C (which needs to be converted to J/g°C for consistent units). The equation will be set up as \(-(1.05 \times 4184) = (1590 \times 0.032 \times 4.18) \times (T_f - 78.2)\), where \(T_f\) is the final temperature. Solving for \(T_f\) will give the final temperature of the platinum after cooling.

Key Concepts

Understanding Specific Heat Capacity

Specific heat capacity is the amount of heat required to change the temperature of a unit mass of a substance by one degree Celsius. It's like asking how "thirsty" a material is for heat! The higher the specific heat capacity, the more heat it takes to change the temperature.

When we look at common substances like water and metals, you might notice they have different specific heat capacities. For example, water has a high specific heat capacity of 4.18 J/g°C, which means it takes more energy to heat it up, compared to something like platinum, which has a specific heat capacity of just 0.032 cal/g°C.

• Water's high specific heat helps it regulate temperature in the environment.
• Metals generally have lower specific heats, which means they heat up and cool down quickly.

This property is crucial in many applications, from cooking to engineering and climate science.

Calculating Temperature Change

Temperature change calculation uses the equation: \[ q = m \cdot c \cdot \Delta T \] Here, \( q \) is the heat added (in joules), \( m \) is the mass (in grams), \( c \) is the specific heat capacity of the material and \( \Delta T \) is the change in temperature.

To find out how much a substance's temperature changes when heat is added or removed, rearrange this formula to solve for the change in temperature, \( \Delta T \).\[ \Delta T = \frac{q}{m \cdot c} \]

• This relationship shows that the more heat you add, the bigger the temperature change, especially if the material has a low specific heat.
• If the material's mass is higher, the temperature change is smaller for the same amount of heat.

This is handy not only in chemistry labs, but also in everyday life, like figuring out how long it takes to cool a hot cup of coffee.

Exploring Calorimetry Concepts

Calorimetry is the science of measuring the heat of chemical reactions or physical changes. It's a powerful tool to understand energy changes in systems. At its core, it's about tracking energy in the form of heat. When a substance undergoes a change and either absorbs or releases heat, calorimetry can help measure that change precisely.

There are key points to understand in calorimetry:

• Heat gained or lost by the substance must equal the heat lost or gained by the surroundings. This is based on the principle of conservation of energy.
• All calculations should use consistent units, converting where necessary, like kcal to Joules.
• Insulation is key in lab-based calorimetry to ensure that no heat escapes.

Understanding calorimetry is essential for fields like chemistry and physics, as well as in fields like nutrition, where it's used to calculate the caloric content of food using the energy derived from combustion.