470 cm3 of water at 80°C is mixed with 145 cm3 of water at 18°C. Calculate the f

Question

470 cm3 of water at 80°C is mixed with 145 cm3 of water at 18°C. Calculate the final equilibrium temperature, assuming no heat is lost to outside the water. Equilibrium temperature:

Imagine now that one tried to repeat the same measurement, but between the time that the 470 cm3 of water was measured at 80°C and the time it was mixed with the 145 cm3 of water at 18°C, the warmer water had lost 3290 cal to the environment. Calculate the final equilibrium temperature assumning no heat loss after mixing occurs. Equilibrium temperature after initial heat loss:

Explanation / Answer

Solution: Part (1)

Volume of hot water Vh = 470cm3 at a temperature Th = 80oC

Volume of cold water Vc = 145cm3 at a temperature Tc = 18oC

Since the density of water is d = 1g/cm3

The mass of hot water is, mh = volume*density = (470cm3)*( 1g/cm3) = 470 g

The mass of cold water is, mc = (145cm3)*( 1g/cm3) = 145 g

Specific heat of water c = 1 cal./g.oC

Let the T be the final equilibrium temperature of the mixture of water after mixing and let m be the mass of the mixture, m = 470g+145g = 615 g.

Since no heat lost outside the system, the energy transfer equation can be written as

heat lost by the hot water = heat gained by the cold water

c*mh*(Th – T) = c*mc*(T – Tc)

cancelling c term from both sides and expanding the bracket

mh*Th - mh*T = mc*T - mc*Tc

mc*T + mh*T = mh*Th + mc*Tc

T*( mc + mh) = mh*Th + mc*Tc

T = [mh*Th + mc*Tc ]/ ( mc + mh)

T = [(470g)*(80oC) + (145)*(18oC)]/(470g + 145g)

T = 65.38oC

Thus the final temperature of the mixture of water is 65.38 oC

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Part (B) Now since the warmer water has lost 3290 cal to the environment before it was mixed with the colder water, we get,

heat lost by the hot water =heat lost to environment + heat gained by the cold water

c*mh*(Th – T) = 3290 cal + c*mc*(T – Tc)

c*mh*Th – c*mh*T = 3290 cal + c*mc*T – c*mc*Tc

c*mh*Th + c*mc*Tc = 3290 cal + c*mc*T + c*mh*T

(1cal/g.oC)*(470g)*(80oC) + (1cal/g.oC)*(145g)*(18oC) = 3290 cal + c*mc*T + c*mh*T

3290 cal + c*mc*T + c*mh*T = 40210 cal

c*mc*T + c*mh*T = 40210 cal - 3290 cal

T*c*(mc + mh) = 36920 cal

T = 36920 cal / c*(mc + mh)

T = 36920 cal /[(1cal/g.oC)*(470g+145g)]

T = 60.03 oC

Hence the final temperature of the mixture will be 60.03 oC in this case

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