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Question: A coffee maker heats \(2\;kg\) of water from \[15\]Celsius to \[100\] Celsius. How much thermal ener...

A coffee maker heats 2  kg2\;kg of water from 1515Celsius to 100100 Celsius. How much thermal energy was required?

Explanation

Solution

We know that thermal energy or the heat energy is the increase in temperature due to the movement of atoms. Using the given formula, since the mass and the temperature difference is given, we can calculate the thermal energy from the for thermodynamics formula.
Formula used:
Q=m×c×ΔTQ=m\times c\times \Delta T

Complete answer:
We know from conservation of energy that energy can neither be created or destroyed but it can be transferred from one form to another. We also know that the molecules in a water being a liquid, can move around freely and hence have a high kinetic energy as compared to that of solids.
When an external stimuli is used to increase the temperature of any substance , here coffee, this increase in temperature, results in an increase in kinetic energy of the water molecules. This change in energy is also called heat energy or more commonly known as thermal energy. This thermal energy is represented as the following, Q=m×c×ΔTQ=m\times c\times \Delta T, where, QQ is the thermal energy, mm is the mass of the substance , cc is the specific heat of the substance, water and ΔT\Delta T is the change in temperature.
Here, given that m=2kgm=2kg and ΔT=10015=85C\Delta T= 100-15=85^{\circ}C
We know, the specific heat of water isc=4.186J/gKc=4.186 J/gK
Substituting the values, in Q=m×c×ΔTQ=m\times c\times \Delta T, we get,
    Q=2kg×4.186kJ/kgC×85C\implies Q=2kg\times 4.186kJ/kgC \times 85 ^{\circ}C
    Q=712kJ\implies Q=712kJ
Thus the thermal energy or heat energy produced due the temperature increase is 712  kJ712\;kJ

Note:
The difference Celsius scale and Kelvin scale are linear. The specific heat cc is a constant which can be found easily on logarithmic tables books. Also note in the above question, we are converting grams to kilograms and joules to kilojoules hence the fraction is still valid.