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How much energy as heat is required to raise the temperature of \(50.00 \mathrm{mL}\) of water from \(25.52^{\circ} \mathrm{C}\) to \(28.75^{\circ} \mathrm{C} ?\) (Density of water at this temperature = \(0.997 \mathrm{g} / \mathrm{mL} .)\)

Short Answer

Expert verified
The heat required is 672.96 J.

Step by step solution

01

Determine the Mass of Water

The density of water is given as 0.997 g/mL, and the volume of water is 50.00 mL. We use the formula for mass: mass = density × volume. So, the mass of water is 0.997 g/mL × 50.00 mL = 49.85 g.
02

Calculate the Temperature Change

To find the change in temperature, we subtract the initial temperature from the final temperature: \( \Delta T = 28.75^{\circ} \mathrm{C} - 25.52^{\circ} \mathrm{C} = 3.23^{\circ} \mathrm{C} \).
03

Use the Specific Heat Formula

The formula for heat energy is \( q = m \times c \times \Delta T \), where:- \( q \) is the heat energy,- \( m \) is the mass (49.85 g),- \( c \) is the specific heat capacity of water (4.18 J/g°C),- \( \Delta T \) is the temperature change (3.23°C).
04

Calculate the Heat Energy

Substitute the known values into the formula: \( q = 49.85 \times 4.18 \times 3.23 \). This gives the heat energy \( q = 672.96 \) J.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Specific Heat Capacity
Specific heat capacity is a key concept in understanding how substances absorb or release energy as heat. It is defined as the amount of heat required to change the temperature of one gram of a substance by one degree Celsius. Each substance has its own specific heat capacity, meaning different materials require different amounts of energy to change temperature.

Here's a practical example: Water has a specific heat capacity of 4.18 J/g°C, which is relatively high compared to other substances. This means it can absorb a lot of heat without a large change in temperature. This is why water is excellent for regulating temperature in environments, such as in oceans and even our bodies.

When calculating energy changes with specific heat, you'll see it in formulas as the variable 'c'. This "c" value is crucial for determining how much heat is absorbed or released. To solve problems involving specific heat, you use the formula \( q = m \times c \times \Delta T \), where \( q \) is the heat energy involved, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the temperature change of the substance.
Temperature Change
Temperature change, denoted \( \Delta T \), is the difference in temperature between the final and initial states of a system. Understanding temperature change is crucial because it helps us determine how much energy is transferred during a heating or cooling process.

In the context of heat energy calculations, you calculate the change in temperature by subtracting the initial temperature \( T_i \) from the final temperature \( T_f \):
  • \( \Delta T = T_f - T_i \)

For example, if you have water that starts at \( 25.52^{\circ}\mathrm{C} \) and is heated to \( 28.75^{\circ}\mathrm{C} \), the temperature change is \( 3.23^{\circ}\mathrm{C} \). This difference is crucial in calculating how much energy is required to achieve this change. The greater the temperature change, the more energy needed, assuming mass and specific heat remain constant.

Always check your initial and final temperatures carefully, since any mistake could lead to incorrect conclusions about the amount of energy required.
Mass Calculations
Mass calculations are an essential part of determining the amount of energy required to change a substance’s temperature. The mass of a substance directly affects the total energy needed, essentially acting as a "multiplier" in energy calculations.

In many exercises, you might be given density and volume, like in our original exercise, where water's density is 0.997 g/mL, and the volume is 50.00 mL. You would calculate the mass using the formula:
  • mass = density × volume
  • mass = 0.997 g/mL × 50.00 mL = 49.85 g

Understanding mass is vital because even a small change in mass can significantly alter the heat calculations. More mass means more substance to heat and thus more energy required. Calculating mass correctly ensures that you're using the right amount of substance in your heat calculations.

Always double-check your mass calculations, especially when converting from volume, to ensure the accuracy of your entire heat calculation.

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Most popular questions from this chapter

You wish to know the enthalpy change for the formation of liquid \(\mathrm{PCl}_{3}\) from the elements. $$ \mathrm{P}_{4}(\mathrm{s})+6 \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{PCl}_{3}(\ell) \quad \Delta_{\mathrm{r}} H^{\circ}=? $$ The enthalpy change for the formation of \(\mathrm{PCl}_{5}\) from the elements can be determined experimentally, as can the enthalpy change for the reaction of \(\mathrm{PCl}_{3}(\ell)\) with more chlorine to give \(\mathrm{PCl}_{5}(\mathrm{s}):\) \(\begin{aligned} \mathrm{P}_{4}(\mathrm{s})+10 \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{PCl}_{5}(\mathrm{s}) & \\ \Delta_{r} H^{\circ} &=-1774.0 \mathrm{kJ} / \mathrm{mol}-\mathrm{rxn} \\\ \mathrm{PCl}_{3}(\ell)+\mathrm{Cl}_{2}(\mathrm{g}) \rightarrow \mathrm{PCl}_{5}(\mathrm{s}) & \\ \Delta_{\mathrm{r}} H^{\circ} &=-123.8 \mathrm{kJ} / \mathrm{mol}-\mathrm{rxn} \end{aligned}\) Use these data to calculate the enthalpy change for the formation of 1.00 mol of \(\mathrm{PCl}_{3}(\ell)\) from phosphorus and chlorine.

You are attending summer school and living in a very old dormitory. The day is oppressively hot, there is no air-conditioner, and you can't open the windows of your room. There is a refrigerator in the room, however. In a stroke of genius, you open the door of the refrigerator, and cool air cascades out. The relief does not last long, though. Soon the refrigerator motor and condenser begin to run, and not long thereafter the room is hotter than it was before. Why did the room warm up?

Determine whether energy as heat is evolved or required, and whether work was done on the system or whether the system does work on the surroundings, in the following processes at constant pressure: (a) Liquid water at \(100^{\circ} \mathrm{C}\) is converted to steam at \(100^{\circ} \mathrm{C}\) (b) Dry ice, \(\mathrm{CO}_{2}(\mathrm{s}),\) sublimes to give \(\mathrm{CO}_{2}(\mathrm{g})\)

When \(108 \mathrm{g}\) of water at a temperature of \(22.5^{\circ} \mathrm{C}\) is mixed with \(65.1 \mathrm{g}\) of water at an unknown temperature, the final temperature of the resulting mixture is \(47.9^{\circ} \mathrm{C} .\) What was the initial temperature of the second sample of water?

The freezing point of mercury is \(-38.8^{\circ} \mathrm{C} .\) What quantity of energy, in joules, is released to the surroundings if \(1.00 \mathrm{mL}\) of mercury is cooled from \(23.0^{\circ} \mathrm{C}\) to -38.8 \(^{\circ} \mathrm{C}\) and then frozen to a solid? (The density of liquid mercury is \(13.6 \mathrm{g} / \mathrm{cm}^{3} .\) Its specific heat capacity is 0.140 J/g \(\cdot\) K and its heat of fusion is \(11.4 \mathrm{J} / \mathrm{g} .\) )

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