Question

In a physics lab, a student accidentally drops a \(25.0-\mathrm{g}\) brass washer into an open dewar of liquid nitrogen at 77.2 K. How much liquid nitrogen boils away as the washer cools from \(293 \mathrm{K}\) to $77.2 \mathrm{K} ?\( The latent heat of vaporization for nitrogen is \)199.1 \mathrm{kJ} / \mathrm{kg}$.

Short answer

Answer: Approximately 10.42 g of liquid nitrogen boils away.

Step-by-step solution

Calculate the heat lost by the brass washer

To calculate the heat transferred from the brass washer to the liquid nitrogen, we will use the formula: \(q = mc\Delta T\) where q is the heat transfer, m is the mass of the brass washer, c is the specific heat capacity of the brass, and \(\Delta T\) is the temperature difference. Given the mass of the brass washer (\(m = 25.0\,\text{g}\)) and the initial and final temperatures (\(T_1 = 293\,\text{K}\), \(T_2 = 77.2\,\text{K}\)), we need to find the specific heat capacity of brass (c). From reference tables, the specific heat capacity of brass is approximately \(c = 0.385\,\mathrm{J/(g\cdot K)}\). The temperature difference is \(\Delta T = T_1 - T_2 =293\,\text{K} - 77.2\,\text{K}\). Now we can calculate the heat loss: \(q = mc\Delta T\)

Calculate the temperature difference and heat loss

Calculate the temperature difference: \(\Delta T = 293\,\text{K} - 77.2\,\text{K} = 215.8\,\text{K}\) Now, calculate the heat loss: \(q = (25.0\,\mathrm{g}) (0.385\,\mathrm{J/(g\cdot K)}) (215.8\,\mathrm{K})\) \(q \approx 2075.45\,\text{J}\)

Calculate the amount of liquid nitrogen that evaporates

Now we will use the heat loss of the brass washer (\(q = 2075.45\,\text{J}\)) and the latent heat of vaporization for liquid nitrogen (\(L_V = 199.1\,\text{kJ/kg}\)) to calculate the mass of the liquid nitrogen that evaporates (m). The formula to calculate the mass of the liquid nitrogen is: \(m = \frac{q}{L_V}\) But first, we need to convert the latent heat of vaporization from \(\mathrm{kJ/kg}\) to \(\mathrm{J/g}\): \(199.1\,\mathrm{kJ/kg} \times \frac{1000\,\mathrm{J}}{1\,\mathrm{kJ}} \times \frac{1\,\mathrm{kg}}{1000\,\mathrm{g}} = 0.1991\,\mathrm{J/g}\) Now, calculate the mass of the liquid nitrogen that evaporates: \(m = \frac{2075.45\,\text{J}}{0.1991\,\mathrm{J/g}}\) \(m \approx 10.42\,\text{g}\)

Present the final answer

The amount of liquid nitrogen that boils away as the brass washer cools from \(293\,\text{K}\) to \(77.2\,\text{K}\) is approximately \(10.42\,\text{g}\).