Question

A sample of nickel is heated to \(99.8^{\circ} \mathrm{C}\) and placed in a coffeecup calorimeter containing 150.0 \(\mathrm{g}\) water at \(23.5^{\circ} \mathrm{C}\) . After the metal cools, the final temperature of metal and water mixture is \(25.0^{\circ} \mathrm{C}\) . If the specific heat capacity of nickel is 0.444 \(\mathrm{J} /^{\prime} \mathrm{C} \cdot \mathrm{g}\) what mass of nickel was originally heated? Assume no heat loss to the surroundings.

Short answer

The mass of the nickel sample that was originally heated is approximately \(33.1\,g\).

Step-by-step solution

Identify the known and unknown values

: We are given: - Initial temperature of nickel, \(T_{Ni_{initial}} = 99.8^{\circ} \mathrm{C}\) - Initial temperature of water, \(T_{water_{initial}} = 23.5^{\circ} \mathrm{C}\) - Final temperature of metal-water mixture, \(T_{final} = 25.0^{\circ} \mathrm{C}\) - Mass of water present, \(m_{water} = 150.0\,g\) - Specific heat capacity of nickel, \(C_{Ni} = 0.444\, \frac{\mathrm{J}}{^{\circ} \mathrm{C} \cdot \mathrm{g}}\) We need to find the mass of the nickel, \(m_{Ni}\).

Write the heat transfer equation

: Since there is no heat loss to the surroundings, the heat gained by water (\(Q_{water}\)) equals the heat lost by nickel (\(Q_{Ni}\)). We can write the heat transfer equation as follows: \(Q_{water} = -Q_{Ni}\)

Calculate the heat gained by water

: The heat gained by water can be calculated using the specific heat capacity of water, the mass of water, and the temperature difference: \(Q_{water} = m_{water} \cdot C_{water} \cdot (T_{final} - T_{water_{initial}})\) Here, the specific heat capacity of water, \(C_{water} = 4.18\, \frac{\mathrm{J}}{^{\circ} \mathrm{C} \cdot \mathrm{g}}\). Now, we plug in the given values: \(Q_{water} = (150.0\,g) \cdot (4.18\, \frac{\mathrm{J}}{^{\circ} \mathrm{C} \cdot \mathrm{g}}) \cdot (25.0^{\circ} \mathrm{C} - 23.5^{\circ} \mathrm{C})\)

Calculate the heat lost by nickel

: The heat lost by nickel can be calculated using the specific heat capacity of nickel, the mass of nickel, and the temperature difference: \(Q_{Ni} = m_{Ni} \cdot C_{Ni} \cdot (T_{Ni_{initial}} - T_{final})\) Now, we use the equation from Step 2 to substitute for \(Q_{Ni}\): \(Q_{water} = -m_{Ni} \cdot C_{Ni} \cdot (T_{Ni_{initial}} - T_{final})\)

Solve for the mass of nickel

: Now, we can plug in the values of \(Q_{water}\) and \(C_{Ni}\) along with the temperatures to solve for \(m_{Ni}\): \((150.0\,g) \cdot (4.18\, \frac{\mathrm{J}}{^{\circ} \mathrm{C} \cdot \mathrm{g}}) \cdot (25.0^{\circ} \mathrm{C} - 23.5^{\circ} \mathrm{C}) = -m_{Ni} \cdot (0.444\, \frac{\mathrm{J}}{^{\circ} \mathrm{C} \cdot \mathrm{g}}) \cdot (99.8^{\circ} \mathrm{C} - 25.0^{\circ} \mathrm{C})\) Divide both sides by the product of the specific heat capacity of nickel and the temperature difference to find the mass of nickel: \(m_{Ni} = \frac{(150.0\,g) \cdot (4.18\, \frac{\mathrm{J}}{^{\circ} \mathrm{C} \cdot \mathrm{g}}) \cdot (25.0^{\circ} \mathrm{C} - 23.5^{\circ} \mathrm{C})}{(0.444\, \frac{\mathrm{J}}{^{\circ} \mathrm{C} \cdot \mathrm{g}}) \cdot (99.8^{\circ} \mathrm{C} - 25.0^{\circ} \mathrm{C})}\)

Calculate the mass of nickel

: Perform the calculations to determine the mass of nickel: \(m_{Ni} \approx 33.1\,g\) The mass of the nickel sample that was originally heated is approximately \(33.1\,g\).

Key concepts

Specific Heat Capacity

Specific heat capacity is a property of materials that expresses the amount of heat required to change the temperature of a unit mass by one degree Celsius. This concept is crucial in calorimetry, which involves measuring the heat of chemical reactions or physical changes.

In this exercise, we are given the specific heat capacity of nickel, which is necessary to calculate how much heat the nickel loses when it cools down.

This property varies between different materials. For instance, water has a high specific heat capacity, meaning it can absorb a lot of heat without a large change in temperature. Nickel, with its specific heat capacity of 0.444 J/°C·g, absorbs or releases less heat compared to water.

• Water's high specific heat capacity allows it to stabilize temperatures in environments, acting as a heat reservoir.
• When two substances are mixed, the one with lower specific heat capacity often experiences a more significant temperature change.

The formulas involving specific heat capacity, such as in this problem, use the relation:

\[ Q = m \cdot C \cdot \Delta T \] where \(Q\) is the heat transferred, \(m\) is the mass, \(C\) is the specific heat capacity, and \(\Delta T\) is the change in temperature.

Heat Transfer

Heat transfer is the movement of thermal energy from a hotter substance to a cooler one. This concept explains the mixing of nickel and water in the coffee-cup calorimeter. Here, nickel transfers its inner heat to water until both substances reach the same temperature.

In our setup, when the heated nickel is submerged in water, the heat lost by nickel is equal to the heat gained by water until thermal equilibrium is reached.

• Heat always flows from the object at a higher temperature to one at a lower temperature.
• No heat is lost to the surroundings in a perfect calorimeter setup, hence the heat balance formula:
  \[ Q_{lost} = Q_{gained} \]

For this exercise, the heat lost by nickel is calculated by knowing its temperature change and specific heat capacity. Conversely, the heat gained by water can be determined similarly using its specific heat capacity.

By setting these heat amounts equal, and accounting for any negative signs that show directional heat flow, we solve for unknown variables like mass, contributing to the solution of such problems.

Thermal Equilibrium

Thermal equilibrium is a state where two substances in contact with each other reach a common temperature. No heat flows between them once this state is achieved. This concept is foundational in problems like our calorimetry exercise.

Initially, the temperatures of nickel and water differ significantly, but after enough time passes while in contact, their temperatures level out. This means both substances cease exchanging heat, reaching what we call thermal equilibrium.

• This concept is used to find unknowns in calorimetry as it ensures that heat transferred is consistent across calculations.
• An important note is that no external heat is exchanged to or from the surroundings in an ideal isolated system like this setup.

In practical problems, reaching thermal equilibrium allows us to relate the heat lost to the heat gained, leveraging equations for specific heat capacity. Ultimately, the final equilibrium temperature combines both substances' contributions, forming the core around which calorimetry problems are solved.