Question

A 50.0 g sample of liquid water at 25.0 degree C is mixed with 29.0 g of water at 45 degree C. The final temperature of the water is ___________ degree C. The pecific heat capacity of liquid water is 4.18 J/g-K.
A) 32.3
B) 27.6
C) 102
D) 35.0
E) 142

Answer 1

The final temperature of the water is 88.10°C.

Given the following data:

• Initial temperature of liquid water = 25.0°C

• Final temperature of water = 45.0°C

• Mass of liquid water = 50.0 grams

• Mass of water = 29.0 grams

• Specific heat capacity of water = 4.18 J/g°C

To find the final temperature of the water:

The quantity of heat lost by the liquid water = The quantity of heat gained by the water.

$Q_{lost} = Q_{gained}\\\\mc\theta = mc\theta\\\\50(4.18)(25) = 29(4.18)(x - 45)\\\\209(25) = 121.22(x - 45)\\\\5225 = 121.22x - 5454.9\\\\121.22x = 5454.9 + 5225\\\\121.22x = 10679.9\\\\x = \frac{10679.9}{121.22}$

Final temperature = 88.10°C

Therefore, the final temperature of the water is 88.10°C.

Read more: https://brainly.com/question/188778

Answer 2

The final temperature of the water will be $\rm \bold{ 32.3^oC}$.

 

The specific heat capacity formula,

$\rm \bold { Q = mc \Delta T}$

As we know,

heat lost by the water = heat gained by the water

$\rm \bold{ Q_1 = Q_2}\\\rm \bold{ (50)(4.18 ) (x - 25) = ( 29)(4.18) (45 - x ) }$

Solve the equation for x

x = 32.34

Hence, we can conclude that the final temperature of the water will be $\rm \bold{ 32.3^oC}$.

To know more about specific heat capacity, refer to the link:

https://brainly.com/question/11194034?referrer=searchResults