A.) 320 cm^3 of water at 87°C is mixed with 115 cm^3 of water at 4°C. Calculate the final equilibrium temperature, assuming no heat is lost to outside the water. B.) Imagine now that one tried to repeat the same measurement, but between the time that the warm water was measured and the time it was mixed with the cooler water, the warmer water had lost 1850 cal to the environment. Calculate the final equilibrium temperature assuming no additional heat loss after mixing occurs.

A) 62.94 C

B) 58.69 C

Explanation:

Specific heat of water = 1 cal/g/deg C

Density of water = 1 g/cm^3

Let T = final temperature

A)

p*V_1 * c_p * dT_1 = p*V_2 * c_p * dT_2

(87 - T) * 320 = (4 + T) * 115

27840 - 320*T = 460 + 115*T

T = 27380 / 435 = 62.94 C

Answer: Equilibrium temperature with no heat loss is 62.94 C

B)

1850 cal will be released when 320 cm^3 cools by

320 cm^3 = 320 g = 320 cal/deg C

1850 cal means the water cools by 1850 / 320 = 5.78125 deg C

So

(87 - 5.78125 - T) * 320 = (4 + T) * 115

T = 25530 / 435 = 58.69 C

Answer: Equilibrium temperature with heat loss is 58.69 C