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

Answer: The final temperature of water is 32.3°C

Explanation:

When two solutions are mixed, the amount of heat released by solution 1 (liquid water) will be equal to the amount of heat absorbed by solution 2 (liquid 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 solution 1 (liquid water) = 50.0 g

$m_2$ = mass of solution 2 (liquid water) = 29.0 g

$T_{final}$ = final temperature = ?

$T_1$ = initial temperature of solution 1 = 25°C  = [273 + 25] = 298 K

$T_2$ = initial temperature of solution 2 = 45°C  = [273 + 45] = 318 K

c = specific heat of water= 4.18 J/g.K

Putting values in equation 1, we get:

$50.0\times 4.18\times (T_{final}-298)=-[29.0\times 4.18\times (T_{final}-318)]\\\\T_{final}=305.3K$

Converting this into degree Celsius, we use the conversion factor:

$T(K)=T(^oC)+273$

$305.3=T(^oC)+273\\T(^oC)=(305.3-273)=32.3^oC$

Hence, the final temperature of water is 32.3°C