Respuesta :
Here,
Heat gain by the first sample of water + Heat lost by the second sample of water is equal to zero (0).
Now, Mass of water sample one = 108 g (given)
Mass of water sample two = 66.9 g (given)
Temperature for water sample one = [tex]23.9^{0}C[/tex]
Let, temperature for water sample two =x
And, final temperature = [tex]47.2^{0}C[/tex]
Now,
[tex]mass of water sample one\times specific heat of water\times (T_{f} - T_{i}) + mass water sample two\times specific heat of water\times (T_{f} - T_{i}) = 0[/tex]
where, [tex]T_{f}[/tex] = final temperature
[tex]T_{i}[/tex] = initial temperature
Substitute all the given values in above formula:
[tex](108 g\times 4.184 J/g . K\times ( 47.2^{0}C- 23.9^{0}C) )+ (66.9 g \times 4.184 J/g . K\times (47.2^{0}C -x))= 0[/tex]
[tex](451.872 J/K \times (23.3^{0}C)) + (279.9096 J/K\times (47.2^{0}C -x)) = 0[/tex]
[tex](10528.6176 J/K(^{0}C)+ (13211.73312 J/K(^{0}C) -279.9096 J/K\times x)= 0[/tex]
[tex](23740.35072 J/K(^{0}C) -279.9096 J/K\times x)= 0[/tex]
[tex] -279.9096 J/K\times x= -23740.35072 J/K(^{0}C)[/tex]
[tex]x =\frac{23740.35072 J/K(^{0}C)}{279.9096 J/K} [/tex]
[tex x =84.81^{0}C [/tex]
