Question

A 50.0-g sample of liquid water at 25.0 °c is mixed with 35.0 g of water at 51.0 °c. The final temperature of the water is ________ °c.

Answer 1

Answer: The final temperature of the water is 6.28°C

Explanation:

When liquid water is mixed with water, the amount of heat released by liquid water will be equal to the amount of heat absorbed by water.

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)]$       ......(1)

where,

q = heat absorbed or released

$m_1$ = mass of liquid water = 50.0 g

$m_2$ = mass of water = 35.0 g

$T_{final}$ = final temperature = ?°C

$T_1$ = initial temperature of liquid water = 25.0°C

$T_2$ = initial temperature of water = 51.0°C

$c$ = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

$50.0\times 4.186\times (T_{final}-25)=-[35.0\times 4.186\times (T_{final}-51)]$

$T_{final}=6.28^oC$

Hence, the final temperature of the water is 6.28°C

Answer 2

[text]Solution:\\ Heat blance is gained by 50g =Heat balance lost by 25g of water\\ So,\\ \(M1 \times temp + M2 \times temp/M1+M2)\\\\\\\frac{(50.0g\times25.0^{.}c)+(23.0g\times57.0^{.}c)}{50.0g+23.0g}         =35.1^{.}c}[/tex]