A 62.5-g piece of gold at 650. K is dropped into 165 g of H2O (l) at 298 K in an insulated container at 1 bar pressure. Calculate the temperature of the system once equilibrium has been reached. Assume that CP,m for Au and H2O are constant at their values for 298 K throughout the temperature range of interest.

Respuesta :

Answer: [tex]T = 302.060 K[/tex]

Explanation:

The gold piece is cooled by water, meaning that gold is heating the water in the meantime, such that thermal equilibrium is reached. By First Law of Thermodynamics the following relation is found:

[tex]\dot Q_{net,out,gold} = \dot Q_{net,in,water}\\ m_{gold} \cdot c_{p, gold} \cdot (650 K - T) = m_{water} \cdot c_{p, water} \cdot (T - 298 K)[/tex]

Specific heats for gold and water are [tex]129 \frac{J}{kg \cdot K}[/tex] and [tex]4186 \frac{J}{kg \dot K}[/tex]. Last expression is simplified by substituting known variables:

[tex]5240 J - 8.063 \cdot T = 690.690 \cdot T - 205825.62 J\\698.753 \cdot T = 211065.62 J\\T = 302.060 K[/tex]