Question

An amount of water of mass 20g at 0 \[{}^\circ C\] is mixed with 40g of water at 10 \[{}^\circ C\], then the final temperature of the mixture is:

Answer 1

Hint:
This is the problem in which one object at some temperature is mixed with another object and we need to find the equilibrium temperature. We can use here the principle of calorimetry to solve this problem. The whole mixture will wind up at the same temperature.

Complete step by step solution:
Mass of water, \[{{m}_{1}}\]= 20 g
The temperature of this mass of water, \[{{T}_{1}}\]= 0 \[{}^\circ C\]
Mass of water, \[{{m}_{2}}\]= 40 g
The temperature of this mass of water, \[{{T}_{2}}\]= 10 \[{}^\circ C\]
Using the principle of calorimetry, equilibrium temperature is given by the formula \[T=\frac{{{m}_{1}}{{T}_{1}}+{{m}_{2}}{{T}_{2}}}{{{m}_{1}}+{{m}_{2}}}\]
Putting the values, \[T=\frac{0+400}{60}=6.6{}^\circ C\]

Thus, the final temperature of the system is 6.6 \[{}^\circ C\]

Additional Information:
The temperature determines the heat energy possessed by the body. Temperature is the measure of hotness or coldness of the body and heat tends to move from high temperature to low temperature. The reverse of this cannot happen on its own. According to the second law of thermodynamics for heat to move from low temperature to high-temperature work is needed to be done on the system.

Note:
In this problem, we have not converted the given mass from units of g to kg because both in numerator and denominator, they cancel out each other. Also, it was not mentioned the units of temperature so we didn’t convert the temperature from degree Celsius to degree kelvin.