Question

If you combine 230.0 mL of water at 25.00 ∘ C and 140.0 mL of water at 95.00 ∘ C, what is the final temperature of the mixture? Use 1.00 g/mL as the density of water.

Answer 1

Answer: The final temperature of the mixture is 51.49°C

Explanation:

When two samples of water are mixed, the heat released by the water at high temperature will be equal to the amount of heat absorbed by water at low temperature

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

The equation used to calculate heat released or absorbed follows:

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

$m_1\times c\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$       ......(1)

where,

q = heat absorbed or released

$m_1$ = mass of water at high temperature = 140 g     (Density of water = 1.00 g/mL)

$m_2$ = mass of water at low temperature = 230 g

$T_{final}$ = final temperature = ?°C

$T_1$ = initial temperature of water at high temperature = 95.00°C

$T_2$ = initial temperature of water at low temperature = 25.00°C

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

Putting values in equation 1, we get:

$140\times 4.186\times (T_{final}-95)=-[230\times 4.186\times (T_{final}-25)]$

$T_{final}=51.49^oC$

Hence, the final temperature of the mixture is 51.49°C