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Chapter 6: Problem 54

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

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

Step by step solution

01

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
02

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)
03

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.
04

Report the result.

The mass of the nickel sample is approximately 28.34 grams.

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