Question

370 cm3 of water at 80°C is mixed with 120 cm3 of water at 4°C. Calculate the final equilibrium temperature, assuming no heat is lost to outside the water.

Answer 1

Answer : The final temperature of the mixture is $61.4^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

And as we know that,

Mass = Density × Volume

Thus, the formula becomes,

$(\rho_1\times V_1)\times c_1\times (T_f-T_1)=-(\rho_2\times V_2)\times c_2\times (T_f-T_2)$

where,

$c_1$ = $c_2$ = specific heat of water = same

$m_1$ = $m_2$ = mass of water  =  same

$\rho_1$ = $\rho_2$ = density of water = 1.0 g/mL

$V_1$ = volume of water at $80.0^oC$  = $370cm^3=370mL$

$V_2$ = volume of water at $4^oC$  = $120cm^3=120mL$

$T_f$ = final temperature of mixture = ?

$T_1$ = initial temperature of water = $80.0^oC$

$T_2$ = initial temperature of water = $4^oC$

Now put all the given values in the above formula, we get:

$(\rho_1\times V_1)\times (T_f-T_1)=-(\rho_2\times V_2)\times (T_f-T_2)$

$(1.0g/mL\times 370mL)\times (T_f-80.0)^oC=-(1.0g/mL\times 120mL)\times (T_f-4)^oC$

$T_f=61.4^oC$

Therefore, the final temperature of the mixture is $61.4^oC$