Question

A 75 g piece of gold (Au) at 1000 K is dropped into 200 g of H2O at 300K in an insulatedcontainer at 1 bar. Calculate the temperature of the system once the equilibrium has beenreached. Assume that CP,mfor Au and H2O are constant and its value for 298 K throughoutthe temperature range of interest.

Answer 1

the final temperature is T final = 308 K

Step-by-step explanation:

since all heat released by gold is absorbed by water

Q gold + Q water = Q surroundings =0 (insulated)

Assuming first that no evaporation of water occurs , and denoting g as gold and w as water , then

Q gold = m g*cp g* ( T final - T initial g)

Q gold = m w*cp w* ( T final - T initial w)

where

m= mass

cp = specific heat capacity

T final = final temperature

T initial g and T initial w = initial temperature of gold and water respectively

thus

Q gold + Q water = 0

m g*cp g* ( T final - T initial g) + m w*cp w* ( T final - T initial w) =0

m g*cp g* T final + m w*cp w* T final = m g*cp g* T initial g+ m w*cp w* T initial w

T final = (m g*cp g* T initial g+ m w*cp w* T initial w)/(m g*cp g+ m w*cp w)

replacing values and assuming cp w = 1 cal/gK = 4.816 J/gK and cp g = 0.129 J/gK (from tables), then

T final = (75 g*0.129 J/gK* 1000 K + 200 g * 4.816 J/gK * 300 K )/(75 g*0.129 J/gK*+ 200 g * 4.816 J/gK ) = 308 K

T final = 308 K

since T boiling water = 373 K and T final = 308 K , we confirm that water does not evaporate

therefore the final temperature is T final = 308 K