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 \(0.444 \mathrm{J} /^{\circ} \mathrm{C} \cdot \mathrm{g}\) what mass of nickel was originally heated? Assume no heat loss to the surroundings.

We can calculate the heat gained by water using the following formula: \[Q_{water} = m_{water} \times c_{water} \times \Delta T_{water}\] where: \(Q_{water}\) = Heat gained by water (Joules) \(m_{water}\) = mass of water = 150 g \(c_{water}\) = specific heat capacity of water = 4.18 J/g°C \(\Delta T_{water}\) = change in temperature of water = 25.0 °C - 23.5 °C. Calculate \(\Delta T_{water}\): \[\Delta T_{water} = 25.0 - 23.5 = 1.5^{\circ}\mathrm{C}\] Now, calculate \(Q_{water}\): \[Q_{water} = 150 \times 4.18 \times 1.5 = 940.5\ \mathrm{J}\]

Since there is no heat loss to the surroundings, the heat gained by water should be equal to the heat lost by the nickel. Thus, \[Q_{nickel} = 940.5\ \mathrm{J}\]

We can calculate the mass of nickel using the following formula: \[Q_{nickel} = m_{nickel} \times c_{nickel} \times \Delta T_{nickel}\] where: \(Q_{nickel}\) = Heat lost by nickel (Joules) = 940.5 J \(m_{nickel}\) = mass of nickel (grams) \(c_{nickel}\) = specific heat capacity of nickel = 0.444 J/g°C \(\Delta T_{nickel}\) = change in temperature of nickel = 99.8 °C - 25.0 °C. Calculate \(\Delta T_{nickel}\): \[\Delta T_{nickel} = 99.8 - 25.0 = 74.8^{\circ}\mathrm{C}\] Now, calculate \(m_{nickel}\): \[m_{nickel} = \frac{Q_{nickel}}{c_{nickel} \times \Delta T_{nickel}} = \frac{940.5}{0.444 \times 74.8} \approx 28.04 \ \mathrm{g}\] Thus, the mass of nickel that was originally heated is approximately 28.04 grams.