
Calculate the amount of heat required to convert $ 45\,g $ of water at $ {43^o}\,C $ to stream at $ {100^o}\;C $ $ ? $
Answer
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Hint: First the given water must be heated to $ {100^o}\;C $ . Then the $ {100^o} $ water must be turned into steam. Using the equation $ Q = mc\Delta T $ we can calculate the amount of energy for heating the water to $ {100^o}\;C $ . Where $ Q $ is energy input, $ m $ is mass of the matter to heat, $ c $ is specific heat capacity of the matter to heat and $ \Delta T $ is the temperature change of the matter.
Complete Step By Step Answer:
Heat is the form of energy that is transferred between systems or objects with different temperatures. Heat is generally measured in joules ( $ J $ ).
The amount of energy calculated by using formula
$ Q = mc\Delta T $ -----(1)
Given mass of water $ m = 45\,g $ , initial temperature $ {T_i} = {43^o}\,C $ , and final temperature $ {T_f} = {100^o}\,C $ .
Then $ \Delta T = {T_f} - {T_i} = {57^o}\,C $
Also, we know that $ c = 4187\,J\,K{g^{ - 1}}{K^1} $ (the specific heat capacity of water)
Then the equation (1) becomes
$ Q = \left( {0.045} \right) \times \left( {4187} \right) \times \left( {57} \right) $
$ \Rightarrow Q = 10721 \approx 10.7\,kJ $
Now we must turn the water into steam. This is done by increasing the potential energy of the water without changing its temperature using something called Specific latent heat of evaporation of water. It is given by the equation
$ Q = mL $ ------(2)
Where, $ L $ is the specific latent heat
We know that the latent heat of evaporation of water
$ L = 2264.705\,kJ/kg $
Plugging it in to the equation:
$ Q = 2264.705 \times 0.045 \approx 101.9\,kJ $
So, the total energy change $ = 101.9 + 10.7 \approx 112.6\,kJ $
Hence the amount of heat required to convert $ 45\,g $ of water at $ {43^o}\,C $ to stream at $ {100^o}\;C $ is $ 112.6\,kJ $ .
Note:
Heat has no shape, no mass, no colour, no odour, no volume. Also note that heat flowing from the high-temperature system to the low-temperature system. Heat energy is also known as thermal energy.
Specific latent heat of evaporation of water is the energy amount required to convert a unit mass of water into steam at constant temperature.
Complete Step By Step Answer:
Heat is the form of energy that is transferred between systems or objects with different temperatures. Heat is generally measured in joules ( $ J $ ).
The amount of energy calculated by using formula
$ Q = mc\Delta T $ -----(1)
Given mass of water $ m = 45\,g $ , initial temperature $ {T_i} = {43^o}\,C $ , and final temperature $ {T_f} = {100^o}\,C $ .
Then $ \Delta T = {T_f} - {T_i} = {57^o}\,C $
Also, we know that $ c = 4187\,J\,K{g^{ - 1}}{K^1} $ (the specific heat capacity of water)
Then the equation (1) becomes
$ Q = \left( {0.045} \right) \times \left( {4187} \right) \times \left( {57} \right) $
$ \Rightarrow Q = 10721 \approx 10.7\,kJ $
Now we must turn the water into steam. This is done by increasing the potential energy of the water without changing its temperature using something called Specific latent heat of evaporation of water. It is given by the equation
$ Q = mL $ ------(2)
Where, $ L $ is the specific latent heat
We know that the latent heat of evaporation of water
$ L = 2264.705\,kJ/kg $
Plugging it in to the equation:
$ Q = 2264.705 \times 0.045 \approx 101.9\,kJ $
So, the total energy change $ = 101.9 + 10.7 \approx 112.6\,kJ $
Hence the amount of heat required to convert $ 45\,g $ of water at $ {43^o}\,C $ to stream at $ {100^o}\;C $ is $ 112.6\,kJ $ .
Note:
Heat has no shape, no mass, no colour, no odour, no volume. Also note that heat flowing from the high-temperature system to the low-temperature system. Heat energy is also known as thermal energy.
Specific latent heat of evaporation of water is the energy amount required to convert a unit mass of water into steam at constant temperature.
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