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# Calorimetry Formula

Calorimetry is the part of chemistry which is about the study of the quantity of heat which is absorbed or released with the surrounding during some chemical reaction. We measure it using equipment, calorimeter. It has a thermometer and measures the variable called has eaten of combustions. This topic will help students to understand the concept of calorimeter with Calorimetry Formula with examples. Let us learn it!

## Concept of Calorimetry

Calorimetry defines the act of measuring the different changes in the state variables of a body for deriving the heat transfer related to the changes of its states. We can perform this Calorimetry with a calorimeter.

In calorimetry, the value of the heat of combustion is constant for each substance (it is commonly known as the heat of combustion standard) and it is measured with the calorimetry (as discussed above) and this value corresponds to the heat released during the combustion of 1 mole of a certain substance. Source:en.wikipedia.org

A calorimeter is a device that is used to measure the quantity of heat transferred to or from an object. Coffee cup calorimeter is usually filled with water and used as a common calorimeter. The more sophisticated cases are a lid on the cup with an inserted thermometer and maybe even a stirrer.

Chemists often use another one known as a bomb calorimeter. It measures the heat exchanges associated with chemical reactions, especially combustion reactions. Thus, if we need to find out the heat of combustion of 2 g of some substance, it should be applying the equation for finding the heat value.

### The Formula for Calorimetry:

It is as follows: $$Q_{sub} = m \times C \times \Delta T$$

Where,

 $$Q_{sub}$$ Heat Energy of substance m Mass C Specific Heat $$\Delta T$$ Temperature Difference

## Solved Examples for Calorimetry Formula

Q.1: A physics class has been assigned the task to determine an experimental value for the heat of fusion of ice. Student mass out 25.8 grams of ice and place it into a coffee cup with 100.0 g of water at $$35.4 ^{\circ} C$$. Then they place a lid on the coffee cup and insert a thermometer. After a few minutes, the ice has completely melted and the water temperature has lowered to $$18.1^{\circ}C$$. What will be their experimental value for the specific heat of fusion of ice? Use Calorimetry Formula.

Solution: The fundamental step for the solution to this problem is the recognition that the quantity of energy lost by the water when cooling is equal to the quantity of energy required to melt the ice. In equation form, we can state this as:

$$Q_{ice} = – Q_{calorimeter}$$

Here the calorimeter is considered to be the water in the coffee cup. Also, the mass of this water and its temperature change is known, then the value can be determined. Thus,

$$Q_{ice} = m \times C \times \Delta T \\$$

$$Q_{ice} = 100 \times 4.18 \times (18.1 – 35.4) = -7231.4 J \\$$

The negative sign here indicates that the water lost energy. The assumption is that this energy lost by the water will be equal to the quantity of energy gained by the ice. So $$Q_{ice} = +7231.4 J$$. Here, the positive sign indicates an energy gain. We can use this value with the equation to determine the heat of fusion of the ice.

$$Q_{ice} = m_{ice} \times \Delta H_{ice}$$

$$+7231.4 J = (25.8 g) \times H_{ice}$$

Thus, $$\Delta H_{ice} = \frac {7231.4 J} {25.8 g}$$

$$\Delta H_{ice} = 280.28 J per gram.$$

Therefore, the experimental value for the specific heat of fusion of ice is 280.28 Joule per gram.

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