If in 1.1 kg of water which is contained in a calorimeter...

If in $1.1 \mathrm{~kg}$ of water which is contained in a calorimeter of water equivalent $0.02 \mathrm{~kg}$ at $15^{\circ} \mathrm{C}$, steam at $100^{\circ} \mathrm{C}$ is passed, till the temperature of calorimeter and its contents rises to $80^{\circ} \mathrm{C}$. The mass of steam condensed in kilogram is: (a) $0.131$ (b) $0.065$ (c) $0.260$ (d) $0.135$

Key Concepts

Calorimetry

Calorimetry is the science of measuring the amount of heat exchanged in physical processes, such as chemical reactions or phase changes. It involves using a calorimeter, which is designed to absorb or supply known amounts of energy. In problems like this, the water equivalent of the calorimeter is taken into account, meaning that the calorimeter itself behaves as if it were made of a certain mass of water in terms of its heat absorption capacity.

Conservation of Energy in Heat Transfer

The principle of conservation of energy states that energy cannot be created or destroyed, only transferred from one form to another. In this context, the heat lost by the condensing steam (including the latent heat released upon condensation and the sensible heat lost as the condensed steam cools) must equal the heat gained by the existing water and the calorimeter.

Latent Heat of Vaporization

Latent heat of vaporization is the energy released or absorbed by a substance during a phase change between liquid and vapor, without changing its temperature. When steam condenses, it releases a specific amount of energy per unit mass, and this released energy is a crucial part of the heat exchange calculation in calorimetry problems involving phase changes.

Specific Heat Capacity

Specific heat capacity is defined as the amount of heat required to raise the temperature of a unit mass of a substance by one degree Celsius. It is a key property in calculating the heat absorbed or released by a substance as its temperature changes, which is essential in determining the energy balance in calorimetry experiments.

Transcript

00:01 Hello students in this question we have mass of water m w this is equal to 1 .1 kilogram or equal to 100 1 ,100 gram which is contained in the calorimeter of water equivalent m w this is 0 .02 kilogram or we can write here this is 20 gram at the temperature t1 equals to 15 degrees centigrade steam at temperature t2 equals to 100 degree centigrade is passed, okay? and the final temperature t equals to 80 degrees centigrade in the calorie meter.

00:39 So we have to determine the mass of the steam which is condensed.

00:43 So first of all the heat absorbed by the water we can write here.

00:48 So heat absorbed by the water, this is equals to the mass of water, multiplied by specific heat of water, multiplied temperature difference that is final temperature t minus initial temperature t1, plus the heat absorbed by this calorie meter.

01:07 So mw multiplied by this specific heat sw, multiplied by temperature change, which is t t minus t1.

01:16 So heat absorbed by the water will be equals to the mw which is 1100, multiplied by 1, multiplied by t which is 80 minus t1 which is 15 plus 20, multiplied by 1.

01:31 And this 80 minus 15 okay so this can be solved further or we can solve it later okay so let us solve this completely so this is equals to 71500 plus plus this is equal to 1 ,300 okay so this can be solved further so this is equals to 72800 calorie this is the heat absorbed by the water okay now we can calculate the heat lost by the steam so we can write here that heat lost by the steam this is equals to mass of steam multiplied by latent heat of steam plus mass of steam multiplied by the temperature difference okay or the specific heat we can take here okay so we can write specific heat of water and this is temperature difference this is temperature t2 minus the final temperature that is t.

02:39 Okay...

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