Question

If 46.9 mL of ethanol (density = 0.789 g/mL) initially at 7.2°C is mixed with 45.4 mL of water (density = 1.0 g/mL) initially at 28.5°C in an insulated beaker, and assuming that no heat is lost, what is the final temperature of the mixture?

Answer 1

Water has a specific heat of 4.184 J/g C. The specific heat of liquid ethanol, C2H5OH(l), is 2.46 J/g·°C. Let the final temperature be T deg C

Volume is multiplied with density to obtain mass.

The mass of ethanol $= \text { 49.6 mL } \times \text { 0.789 g/mL } = \text { 39.13 g}$

The mass of water $= \text { 45.4 mL } \times \text { 1 g/mL } = \text { 45.4 g}$

Heat gained by ethanol $= \text { mass }\times \text { specific heat } \times \text { temperature change}$

Heat gained by ethanol $= \text { 39.13 }\times \text { 2.46 J/ (g deg C) } \times \text {(T - 7.2)} ^oC$

Heat gained by ethanol $= 96.27 \text {(T - 7.2) J}$

Heat lost by water $= \text { mass }\times \text { specific heat } \times \text { temperature change}$

Heat lost by water $= \text { 45.4 g }\times \text { 4.18 J/ (g deg C) } \times \text {(T - 28.5)} ^oC$

Heat lost by water $= 189.772 \times \text {(T - 28.5) J}$

Heat gained by ethanol = - Heat lost by water

$96.27 \text {(T - 7.2) J}= - 189.772 \times \text {(T - 28.5) J}$

$\text {(T - 7.2) }= - 1.9712 \times \text {(T - 28.5) }$

$\text {(T - 7.2) }= - 1.9712T + \text {56.18 }$

$\text {2.9712T }= + \text {63.38 }$

$\text {T }= + \text {21.33 }$

Hence, the final temperature of the mixture is 21.33 deg C.