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

Expert verified
Answer: Approximately 10.42 g of liquid nitrogen boils away.

Step by step solution

01

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: \(q = mc\Delta T\)
02

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: \(q = (25.0\,\mathrm{g}) (0.385\,\mathrm{J/(g\cdot K)}) (215.8\,\mathrm{K})\) \(q \approx 2075.45\,\text{J}\)
03

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: \(m = \frac{2075.45\,\text{J}}{0.1991\,\mathrm{J/g}}\) \(m \approx 10.42\,\text{g}\)
04

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 \(10.42\,\text{g}\).

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