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.

: 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, \(m_{Ni}\).

: 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: \(Q_{water} = -Q_{Ni}\)

: 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: \(Q_{water} = (150.0\,g) \cdot (4.18\, \frac{\mathrm{J}}{^{\circ} \mathrm{C} \cdot \mathrm{g}}) \cdot (25.0^{\circ} \mathrm{C} - 23.5^{\circ} \mathrm{C})\)

: 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 \(Q_{Ni}\): \(Q_{water} = -m_{Ni} \cdot C_{Ni} \cdot (T_{Ni_{initial}} - T_{final})\)

: 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: \(m_{Ni} = \frac{(150.0\,g) \cdot (4.18\, \frac{\mathrm{J}}{^{\circ} \mathrm{C} \cdot \mathrm{g}}) \cdot (25.0^{\circ} \mathrm{C} - 23.5^{\circ} \mathrm{C})}{(0.444\, \frac{\mathrm{J}}{^{\circ} \mathrm{C} \cdot \mathrm{g}}) \cdot (99.8^{\circ} \mathrm{C} - 25.0^{\circ} \mathrm{C})}\)

: 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 \(33.1\,g\).