Question

If you combine 280.0 mL of water at 25.00 °C and 140.0 mL of water at 95.00 °C, what is the final temperature of the mixture? Use 1.00 g/mL as the density of water.

If a system has 4.50 × 102 kcal of work done to it, and releases 5.00 × 102 kJ of heat into its surroundings, what is the change in internal energy of the system?

Answer 1

a)
m1 = V1*density
      =280 mL * 1 g/mL
      = 280 g
T1 = 25 oC

m2 = V2*density
      =140 mL * 1 g/mL
      = 140 g
T2 = 95 oC

Let final temperature be ToC
use:
Heat lost by m2 = heat gained by m1
m2*C*(95-T) = m1*C*(T-25)
140*(95-T) = 280*(T-25)
13300 - 140*T = 280*T - 7000
T = 48.33 oC
Answer: 48.33 oC

2)
Q = -5.00 *10^2 KJ
W = 4.50*10^2 KCal
     = 4.5*10^2 * 4.184 KJ
     = 1.88*10^3 KJ

delta U = Q + W
      = (-5.00 *10^2) + (1.88*10^3)
      = 1.38*10^3 KJ
Answer: 1.38*10^3 KJ