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How many calories are required to raise the temperature of 225 g of water from $42^\circ C$ to $75^\circ C$?

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Last updated date: 27th Feb 2024
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IVSAT 2024
Answer
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Hint: In this question, heat required is needed to be calculated. For this the formula used is represented as the product of mass, specific heat and change in the temperature. The heat required is calculated in terms of Joules. Here, we need to determine the heat required in terms of calories.

Complete step by step answer:
It is given that the mass of water is 225 g.
The temperature change is from $42^\circ C$ to $75^\circ C$.
Here in this question we need to determine the amount of heat needed to raise the temperature of 225 g of water from $42^\circ C$ to $75^\circ C$.
To determine the heat required, the heat capacity formula is applied. The heat capacity formula is expressed as the product of the mass, specific heat and the change in temperature.
The heat capacity formula is shown below.
$Q = mc\Delta T$
Where,
Q is the heat capacity is Joules
m is the mass in grams
c specific heat capacity of the substance in J/$g^\circ C$.
$\Delta T$ is a change in temperature.
The change in temperature will be $75^\circ C - 42^\circ C = 33^\circ C$
The specific heat capacity of water is 4.2 J/$g^\circ C$.
It means that 4.2 joules of energy is needed to raise 1 gram of water by 1 degree Celsius.
Substitute the values in the above expression.
$ \Rightarrow Q = 225 \times 4.2 \times 33$
$ \Rightarrow Q = 31185$J or 7453.393 calories.

Therefore, 7453.393 calories are required to raise the temperature of 225 g of water from $42^\circ C$ to $75^\circ C$

Note: The formula of heat capacity can also be used to calculate the heat capacity, mass or the difference in temperature of the substance. 1 calorie is equal to 4.184 Joules so 1 Joule is equal to 0.239005736 calorie.

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