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

Short answer

The mass of the nickel sample is approximately \(28.34\) grams.

Step-by-step solution

Write the heat transfer equation for nickel and water.

Using the formula for heat transfer (Q = mcΔT), we write the heat transfer equations for nickel and water: Q_nickel = m_nickel * c_nickel * (T_final - T_initial_nickel) Q_water = m_water * c_water * (T_final - T_initial_water) Since there is no loss of heat to the surroundings, the heat gained by the water would be equal to the heat lost by the nickel: Q_nickel = Q_water

Plug in the known values.

We know the following: c_nickel = 0.444 J/°C·g (specific heat capacity of nickel) m_water = 150.0 g c_water = 4.18 J/°C·g (specific heat capacity of water) T_final = 25.0 °C T_initial_nickel = 99.8 °C T_initial_water = 23.5 °C We can plug these values into the equations from Step 1: m_nickel * 0.444 J/°C·g * (25.0 °C - 99.8 °C) = 150.0 g * 4.18 J/°C·g * (25.0 °C - 23.5 °C)

Solve for m_nickel (mass of nickel).

Solve for m_nickel in the equation: m_nickel * 0.444 J/°C·g * (-74.8 °C) = 150.0 g * 4.18 J/°C·g * (1.5 °C) m_nickel * (-33.1912 J/g) = 940.5 J Now divide both sides by (-33.1912 J/g): m_nickel = 940.5 J / (-33.1912 J/g) m_nickel = -28.34 g (approximately) It doesn't make sense to have a negative mass value, but the negative sign comes from the heat transfer direction (heat lost by nickel and heat gained by water). Thus, we can report the mass as a positive value.

Report the result.

The mass of the nickel sample is approximately 28.34 grams.