What would be the final temperature of a mixture of $50g$ of water at ${20^0}C$ temperature and $50g$ of water at ${40^0}C$?
Answer
650.4k+ views
Hint
We will use the concept of calorimeter. We can use that Amount of heat radiated by a hot object equals the amount of heat absorbed by the same object at a relatively lower temperature.
$\Rightarrow Amount{\text{ }}of{\text{ }}absorption{\text{ }}or{\text{ }}radiation{\text{ }}of{\text{ }}heat = m \times s \times T$
Where, $m$ is the mass of the object, $s$ is its specific heat capacity and $T$ is the temperature at which it is placed.
Complete step by step answer
Let the specific heat capacity of water be ${s_w}$
Given, Mass of water used for both cases, ${m_w} = 50g$
Warmer temperature, ${T_1} = {40^0}C$
Cooler Temperature, ${T_2} = {20^0}C$
Now,
$Amount{\text{ }}of{\text{ }}heat{\text{ }}radiated{\text{ }}by{\text{ }}warm{\text{ }}water = {\text{ }}m \times {s_w} \times ({T_1} - T) = 50 \times {s_w} \times (40 - T) = (40 - T) \times {s_w}$
$Amount{\text{ }}of{\text{ }}heat{\text{ absorbed }}by{\text{ cold }}water = {\text{ }}m \times {s_w} \times (T - {T_2}) = 50 \times {s_w} \times (T - 20) = (T - 20) \times {s_w}$
Then,
By the law of calorimeter,
$Amount{\text{ }}of{\text{ }}heat{\text{ }}radiated{\text{ }}by{\text{ }}warm{\text{ }}water = {\text{ }}Amount{\text{ }}of{\text{ }}heat{\text{ }}absorbed{\text{ }}by{\text{ }}cold{\text{ }}water$$ \Rightarrow \left( {40 - T} \right) = \left( {T - 20} \right)$
After further evaluation, we get the value of the solution, $T = {30^0}C$
We can also simply calculate this value,
As, The same substance in this case water at different temperature is mixed,
Thus, the final mixture temperature will turn out to be the average of the individual temperatures. That means,
$T = \left( {{{40}^0}C + {{20}^0}C} \right)/2 = {30^0}C$
Note
The concept of calorimeter is very much essential for the solving of this type of problems. Here, no doubt the value from the calorimeter formula and that after averaging the temperatures are the same, due to the temperatures acting on the same substance. But that is not always the same as there are situations where two different substances are used at different temperatures and we are given to find out the total mixture temperature. Thus, it is more beneficial to use the concept of calorimeter as the chances of uttering an error and getting deviated from the original answer reduces.
We will use the concept of calorimeter. We can use that Amount of heat radiated by a hot object equals the amount of heat absorbed by the same object at a relatively lower temperature.
$\Rightarrow Amount{\text{ }}of{\text{ }}absorption{\text{ }}or{\text{ }}radiation{\text{ }}of{\text{ }}heat = m \times s \times T$
Where, $m$ is the mass of the object, $s$ is its specific heat capacity and $T$ is the temperature at which it is placed.
Complete step by step answer
Let the specific heat capacity of water be ${s_w}$
Given, Mass of water used for both cases, ${m_w} = 50g$
Warmer temperature, ${T_1} = {40^0}C$
Cooler Temperature, ${T_2} = {20^0}C$
Now,
$Amount{\text{ }}of{\text{ }}heat{\text{ }}radiated{\text{ }}by{\text{ }}warm{\text{ }}water = {\text{ }}m \times {s_w} \times ({T_1} - T) = 50 \times {s_w} \times (40 - T) = (40 - T) \times {s_w}$
$Amount{\text{ }}of{\text{ }}heat{\text{ absorbed }}by{\text{ cold }}water = {\text{ }}m \times {s_w} \times (T - {T_2}) = 50 \times {s_w} \times (T - 20) = (T - 20) \times {s_w}$
Then,
By the law of calorimeter,
$Amount{\text{ }}of{\text{ }}heat{\text{ }}radiated{\text{ }}by{\text{ }}warm{\text{ }}water = {\text{ }}Amount{\text{ }}of{\text{ }}heat{\text{ }}absorbed{\text{ }}by{\text{ }}cold{\text{ }}water$$ \Rightarrow \left( {40 - T} \right) = \left( {T - 20} \right)$
After further evaluation, we get the value of the solution, $T = {30^0}C$
We can also simply calculate this value,
As, The same substance in this case water at different temperature is mixed,
Thus, the final mixture temperature will turn out to be the average of the individual temperatures. That means,
$T = \left( {{{40}^0}C + {{20}^0}C} \right)/2 = {30^0}C$
Note
The concept of calorimeter is very much essential for the solving of this type of problems. Here, no doubt the value from the calorimeter formula and that after averaging the temperatures are the same, due to the temperatures acting on the same substance. But that is not always the same as there are situations where two different substances are used at different temperatures and we are given to find out the total mixture temperature. Thus, it is more beneficial to use the concept of calorimeter as the chances of uttering an error and getting deviated from the original answer reduces.
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