91Ó°ÊÓ

In this problem, we are dealing with heat transfer to a gas in a closed, rigid tank. Heat transfer is essentially the movement of thermal energy from one object to another. This can happen through various mechanisms like conduction, convection, or radiation. In our case, the energy being transferred to the hydrogen gas is quantified at a rate of 400 W.
Understanding heat transfer is crucial because it helps us determine the temperature change in the gas. The formula for heat transfer is given by \ \( Q = \text{Q̇} \times t \), where \( Q \) is the total heat transfer, \ \( \text{Q̇} \) is the rate of heat transfer, and \( t \) is the time over which heat is transferred. For this exercise, the total heat transfer was calculated as \ \( Q = 400 \text{ W} \times 3600 \text{ s} = 1,440,000 \text{ J} \).
It's important to keep the units consistent to ensure accuracy. Here, we converted the time from hours to seconds to match the units of power in watts.

Specific heat capacity \ \( (c_v) \) is a measure of how much heat energy is required to raise the temperature of a unit mass of a substance by one degree. For gases, we often use specific heat at constant volume (\ \( c_v \) ) or constant pressure (\ \( c_p \) ).
In this problem, the specific heat at constant volume is crucial. For an ideal gas, this can be calculated using the formula: \ \( c_v = \frac{R}{k-1} \), where \ \( R \) is the specific gas constant, and \ \( k \) is the ratio of specific heats. Given \ \( R = 4124 \text{ J/(kg\cdot K)} \) and \ \( k = 1.405 \), we find:
\ \( c_v = \frac{4124}{1.405-1} = 10,409 \text{ J/(kg\cdot K)} \).
This means it takes 10,409 J of energy to increase the temperature of 1 kg of hydrogen by 1 K at constant volume. Understanding specific heat helps in determining how much the temperature will change when heat is added.

Finally, we analyze how the temperature of the hydrogen changes due to the added heat. We start with the equation relating heat transfer to temperature change: \ \( Q = mc_v \triangle T \), where \ \( m \) is the mass and \ \( \triangle T \) is the change in temperature.
Rearranging this formula, we get: \ \( \triangle T = \frac{Q}{mc_v} \). For our exercise, \ \( m = 10 \text{ kg} \), \ \( Q = 1,440,000 \text{ J} \), and \ \( c_v = 10,409 \text{ J/(kg\cdot K)} \). Substituting these values, we find:
\ \( \triangle T = \frac{1,440,000 \text{ J}}{10 \text{ kg} \times 10,409 \text{ J/(kg \cdot K)}} = 13.83 \text{ K} \).
This change is then added to the initial temperature to find the final temperature: \ \( T_2 = T_1 + \triangle T = 293 \text{ K} + 13.83 \text{ K} = 306.83 \text{ K} \). Converting this back to Celsius, we get approximately
\ 33.68^{\bullet}\text{C} \.
By understanding these steps, you can apply similar methods to other heat transfer problems involving ideal gases.