Question

When 108 g of water at a temperature of 23.9 °c is mixed with 66.9 g of water at an unknown temperature, the final temperature of the resulting mixture is 47.2 °c. what was the initial temperature of the second sample of water? (the specific heat capacity of liquid water is 4.184 j/g ⋅ k.)?

Answer 1

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 = $23.9^{0}C$

Let, temperature for water sample two =x

And, final temperature = $47.2^{0}C$

Now,

$mass of water sample one\times specific heat of water\times (T_{f} - T_{i}) + mass water sample two\times specific heat of water\times (T_{f} - T_{i}) = 0$

where, $T_{f}$ = final temperature

$T_{i}$ = initial temperature

Substitute all the given values in above formula:

$(108 g\times 4.184 J/g . K\times ( 47.2^{0}C- 23.9^{0}C) )+ (66.9 g \times 4.184 J/g . K\times (47.2^{0}C -x))= 0$

$(451.872 J/K \times (23.3^{0}C)) + (279.9096 J/K\times (47.2^{0}C -x)) = 0$

$(10528.6176 J/K(^{0}C)+ (13211.73312 J/K(^{0}C) -279.9096 J/K\times x)= 0$

$(23740.35072 J/K(^{0}C) -279.9096 J/K\times x)= 0$

$-279.9096 J/K\times x= -23740.35072 J/K(^{0}C)$

$x =\frac{23740.35072 J/K(^{0}C)}{279.9096 J/K}$

[tex x =84.81^{0}C [/tex]