100g of stea initial at 100 degree celcius is passed through a calorimeter containing 1kg of water at 20 degree celcius. Assuming no heat loose and negligible heat capacity of the calorimeter, calculate the final temperature of the calorimeter and the content. heat of vaporation of water = 540kcal/kg, specific heat of water = 1,00kcal/kg degree celcius

To calculate the final temperature of the calorimeter and its contents, we need to use the principle of energy conservation. The heat lost by the hot object (steam) is equal to the heat gained by the cooler object (water and calorimeter).

First, let's calculate the heat lost by the steam using the equation:

Heat lost = mass x specific heat x change in temperature

The mass of the steam is given as 100g. The specific heat of steam is not mentioned, but we can assume it to be negligible compared to water, so we can ignore it. The change in temperature of the steam can be calculated as the final temperature of the system minus the initial temperature of the steam (100°C).

Heat lost = 100g x 1 cal/g°C x (final temperature - 100°C) [1 cal = 1 kcal]

Next, let's calculate the heat gained by the water and calorimeter using the equation:

Heat gained = mass x specific heat x change in temperature

The mass of water is given as 1kg (1000g), and the specific heat of water is given as 1 kcal/kg°C. The change in temperature can be calculated as the final temperature of the system minus the initial temperature of the water (20°C).

Heat gained = 1000g x 1 cal/g°C x (final temperature - 20°C)

Since the heat lost by the steam is equal to the heat gained by the water and calorimeter, we can set up the following equation:

Heat lost = Heat gained

100g x 1 cal/g°C x (final temperature - 100°C) = 1000g x 1 cal/g°C x (final temperature - 20°C)

Simplifying the equation:

final temperature - 100°C = 10 x (final temperature - 20°C)

final temperature - 100°C = 10final temperature - 200°C

9final temperature = 100°C

final temperature = 100°C / 9 ≈ 11.11°C

Therefore, the final temperature of the calorimeter and its contents is approximately 11.11°C.

To solve this problem, we can use the principle of energy conservation.

The amount of heat lost by the hot steam is equal to the sum of the amount of heat gained by the water and the calorimeter.

1. Calculate the heat lost by the hot steam:
The heat lost by the steam can be calculated using the formula:

Qlost = mass * specific heat * (final temperature - initial temperature)

Qlost = 100g * 1.00 kcal/kg°C * (final temperature - 100°C)

2. Calculate the heat gained by the water:
The heat gained by the water can be calculated using the formula:

Qgained = mass * specific heat * (final temperature - initial temperature)

Qgained = 1kg * 1.00 kcal/kg°C * (final temperature - 20°C)

3. Calculate the heat gained by the calorimeter:
Since the heat capacity of the calorimeter is negligible, we assume that all the heat gained by the water is transferred to the calorimeter.

Qgained_by_calorimeter = Qgained = 1kg * 1.00 kcal/kg°C * (final temperature - 20°C)

4. Set up the energy conservation equation:
Since there is no heat lost to the surroundings, the energy gained by the water and the calorimeter is equal to the energy lost by the steam.

Qlost = Qgained + Qgained_by_calorimeter

Substituting the values calculated above:

100g * 1.00 kcal/kg°C * (final temperature - 100°C) = 1kg * 1.00 kcal/kg°C * (final temperature - 20°C) + 1kg * 1.00 kcal/kg°C * (final temperature - 20°C)

Simplifying the equation:

0.10 kcal/°C * (final temperature - 100°C) = 1.00 kcal/°C * (final temperature - 20°C) + 1.00 kcal/°C * (final temperature - 20°C)

0.10 kcal/°C * final temperature - 10 kcal = 2.00 kcal/°C * final temperature - 40 kcal

0.10 kcal/°C * final temperature - 1.00 kcal/°C * final temperature = -40 kcal + 10 kcal

-0.90 kcal/°C * final temperature = -30 kcal

Divide both sides by -0.90 kcal/°C:

final temperature = -30 kcal / -0.90 kcal/°C

final temperature = 33.33 °C

So, the final temperature of the calorimeter and the water content will be approximately 33.33 °C.