A piece of metal of mass \(20.0 \mathrm{~g}\) at \(100.0^{\circ} \mathrm{C}\) is placed in a calorimeter containing \(50.7 \mathrm{~g}\) of watcr at \(22.0^{\circ} \mathrm{C}\). The final temperature of the mixture is \(25.7^{\circ} \mathrm{C}\) What is the specific heat capacity of the mctal? Assume that all the energy lost by the metal is gained by the water.

Short Answer

Expert verified
The specific heat capacity of the metal is \(0.903 \textrm{J/g}^\circ\textrm{C}\).

Step by step solution

01

Understand the Conservation of Energy

According to the principle of conservation of energy, the heat lost by the metal will be equal to the heat gained by the water, assuming there is no heat loss to the surroundings.
02

Calculate the Heat Gained by Water

Use the formula for heat transfer, \( Q = m_c \times c_c \times \Delta T \), where \( m_c \) is the mass of the water, \( c_c \) is the specific heat capacity of water (\(4.18 \textrm{J/g}^\circ\textrm{C} \)) and \( \Delta T \) is the change in temperature of the water.
03

Calculate the Heat Lost by Metal

Since the heat gained by the water is equal to the heat lost by the metal, \( Q = m_m \times c_m \times \Delta T_m \), where \( m_m \) is the mass of the metal, \( c_m \) is the specific heat capacity of the metal, and \( \Delta T_m \) is the change in temperature of the metal.
04

Solving for the Specific Heat Capacity of the Metal

Rearrange the equation for the metal to solve for \( c_m \): \( c_m = \frac{Q}{m_m \times \Delta T_m} \). Then, plug in the values for \( Q \) calculated from the water, the mass of the metal and its temperature change to find \( c_m \.\)
05

Calculate the Temperature Changes

For the water, \( \Delta T = T_{final} - T_{initial} = 25.7^\circ\textrm{C} - 22.0^\circ\textrm{C} \). For the metal, \( \Delta T = T_{initial} - T_{final} = 100.0^\circ\textrm{C} - 25.7^\circ\textrm{C} \).

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Key Concepts

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

Heat Transfer
Understanding heat transfer is crucial for solving problems related to temperature changes in substances. Heat transfer occurs when thermal energy is exchanged due to a temperature difference and can occur through conduction, convection, or radiation. In our exercise, the transfer is through direct contact (conduction) between the metal and the water within the calorimeter.

When the hot metal is placed in the cooler water, heat flows from the metal to the water until thermal equilibrium is reached - that is, until both substances are at the same temperature. The amount of heat transferred is directly proportional to the mass, the specific heat capacity of the substance, and the temperature change.
Conservation of Energy
The principle of the conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. In calorimetry, this principle allows us to set up an equation where the heat lost by the hot object must equal the heat gained by the cooler object, assuming no heat is lost to the surroundings.

In our exercise, the energy lost by the hot piece of metal is completely gained by the cooler water. This energy transformation shows conservation of energy at play—whereby the loss of thermal energy from the metal results in an equivalent gain of thermal energy by the water.
Calorimetry
Calorimetry is the science of measuring the heat of chemical reactions or physical changes. This measurement is conducted using a calorimeter, an insulated device that allows for the accurate measurement of heat transfer.

In our example, the calorimeter ensures that the heat from the metal is not lost to the environment but is transferred to the water inside it. To calculate the specific heat capacity of the metal, we use the known mass, specific heat capacity (an intrinsic property), and temperature change of the water, alongside the mass and temperature change of the metal.
Temperature Change
Temperature change, denoted as \( \Delta T \), is the difference between the final and the initial temperatures. It plays a vital role in calculations involving heat transfer because it reflects the extent to which a substance has absorbed or lost thermal energy.

For this exercise, understanding how to correctly calculate temperature change is critical. For the water, it's the final temperature minus the initial temperature. For the metal, it's the initial temperature (when it's first placed in the calorimeter) minus the final temperature. This temperature change is then used to determine the amount of heat transferred and, consequently, the specific heat capacity of the metal.

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Most popular questions from this chapter

A gas in a cylinder was placed in a heater and gained \(7000 \mathrm{~kJ}\) of heat. If the cylinder increased in volume from \(700 \mathrm{~mL}\) to \(1450 \mathrm{~mL}\) against an atmospheric pressure of 750 Torr during this process, what is the change in internal energy of the gas in the cylinder?

Strong sunshine bombards the Earth with about \(1 \mathrm{~kJ} \cdot \mathrm{m}^{-2}\) in \(1 \mathrm{~s}\). Calculate the maximum mass of pure ethanol that can be vaporized in 10 min from a beaker left in strong sunshine, assuming the surface area of the ethanol to be \(50 \mathrm{~cm}^{2}\). Assume all the heat is used for vaporization, not to increase the temperature.

(a) Describe three ways in which you could increase the internal energy of an open system. (b) Which of these methods could you use to increase the internal energy of a closed system? (c) Which, if any, of these methods could you use to increase the internal energy of an isolated system?

For a certain reaction at constant pressure, \(\Delta U=-95 \mathrm{~kJ}\) and \(56 \mathrm{~kJ}\) of expansion work is done by the system. What is \(\Delta H\) for this process?

Determine the reaction enthalpy for the hydrogenation of ethyne to ethane, \(\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g})+\) \(2 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{~g})\), from the following data: enthalpy of combustion of ethyne, \(-1300 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\); enthalpy of combustion of ethane, \(-1560 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\); enthalpy of combustion of hydrogen, \(-286 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\).

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