please check these;

a container of gas is at pressure of 3.7*10^3 Pa. How much work is done by the gas if its volume expands by 1.6 m.

5.92 * 10^5

A 0.10 kg piece of copper at an initial temperature of 95degC is dropped into 0.20 kg of water contained in a 0.28 kg aluminum calorimeter. The water and calorimeter are initially at 15degC. What is the final temperature of the system when it reaches equilibrium?
(Cp of Copper=387J/kg * degC; Cpof Aluminum=899J/kg * degC; Cp of Water=4186J/kg * degC)

someone gave me a very long and complicated formula for this. does anyone know a simple way to solve this problem? thanks for any help

<<a container of gas is at pressure of 3.7*10^3 Pa. How much work is done by the gas if its volume expands by 1.6 m.

5.92 * 10^5 >>

Please limit yourself to one question per post. What are the dimensions of your answer?

The work done depends upon whether the temperature is constant during the process. To keep it constant, heat has to be added. Your answer (in Joules) is correct if the temperature is held constant.

To calculate the work done by a gas as its volume expands, you can use the formula:

Work = Pressure * Change in Volume

In this case, the pressure is given as 3.7 * 10^3 Pa and the change in volume is 1.6 m^3.

So, the work done by the gas can be calculated as:

Work = 3.7 * 10^3 Pa * 1.6 m^3
= 5920 J
= 5.92 * 10^3 J

Therefore, the work done by the gas is 5.92 * 10^3 J.

Regarding the second question, to find the final temperature when copper is dropped into water contained in an aluminum calorimeter, you can apply the principle of energy conservation. The heat gained by the water and aluminum calorimeter is equal to the heat lost by the copper. The formula can be written as:

mcΔT = mcΔT

Where:
- m is the mass of the object
- c is the specific heat capacity of the substance
- ΔT is the change in temperature

In this case, we have:

(0.10 kg)(387 J/kg * °C)(Tf - 15°C) = (0.20 kg)(4186 J/kg * °C)(Tf - 15°C) + (0.28 kg)(899 J/kg * °C)(Tf - 15°C)

By simplifying and rearranging the equation, we can solve for Tf (the final temperature).

While it is a bit complex, this formula takes into account the specific heat capacities of different substances and ensures accurate results. If you prefer a simpler approach, you can use online calculators or software that can solve these types of equations for you.