How do you calculate the specific heat capacity of ice?
Answer
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Hint: Heat capacity is the proportion of heat consumed by a material to the temperature change. It is normally communicated as calories per degree regarding the real amount of material being thought of, generally a mole (the atomic mass in grams). The heat capacity in calories per gram is called specific heat. The meaning of the calorie depends on the specific heat of the water, characterized as one calorie for each degree Celsius.
Complete step-by-step answer:
The specific heat capacity is a number that tells us how much energy (\[JorKJ\]) is required to raise \[1g\] (\[or \, 1kg\]) by \[1K\]. We can use the following equation:
\[E = m.c.\Delta T\]
In this equation m equals the mass of our sample, c is the specific heat of our sample and \[\Delta T\] is our change in temperature.
For the purposes of this I am going to make up some numbers to show you how to calculate specific heat.
Let’s say we add \[1054J\] of energy to a sample of \[100grams\] of ice and its temperature rises from \[243.15K\] to \[253.15K\]. Using the equation shown above we can calculate the specific heat of ice.
\[1054J = 100g.c.(253.15K - 243.15K)\]
Now we rearrange our equation to solve for c by dividing both sides by \[100g\] and \[10K\] (because \[253.15 - 243.15 = 10\]). Giving us the following equation:
\[1054J.10K = 2.108Jg.K\]
Note: At adequately high temperatures, the heat capacity per atom will in general be the equivalent for all components. For metals of higher nuclear weight, this estimate is now a decent one at room temperature, offering to ascend to the law of Dulong and Petit. For different materials, heat capacity and temperature variety rely upon contrasts in energy levels for atoms.
Complete step-by-step answer:
The specific heat capacity is a number that tells us how much energy (\[JorKJ\]) is required to raise \[1g\] (\[or \, 1kg\]) by \[1K\]. We can use the following equation:
\[E = m.c.\Delta T\]
In this equation m equals the mass of our sample, c is the specific heat of our sample and \[\Delta T\] is our change in temperature.
For the purposes of this I am going to make up some numbers to show you how to calculate specific heat.
Let’s say we add \[1054J\] of energy to a sample of \[100grams\] of ice and its temperature rises from \[243.15K\] to \[253.15K\]. Using the equation shown above we can calculate the specific heat of ice.
\[1054J = 100g.c.(253.15K - 243.15K)\]
Now we rearrange our equation to solve for c by dividing both sides by \[100g\] and \[10K\] (because \[253.15 - 243.15 = 10\]). Giving us the following equation:
\[1054J.10K = 2.108Jg.K\]
Note: At adequately high temperatures, the heat capacity per atom will in general be the equivalent for all components. For metals of higher nuclear weight, this estimate is now a decent one at room temperature, offering to ascend to the law of Dulong and Petit. For different materials, heat capacity and temperature variety rely upon contrasts in energy levels for atoms.
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