Given:
Ma = 31.1 g, the mass of gold
Ta = 69.3 °C, the initial temperature of gold
Mw = 64.2 g, the mass of water
Tw = 27.8 °C, the initial temperature of water
Because the container is insulated, no heat is lost to the surroundings.
Let T °C be the final temperature.
From tables, obtain
Ca = 0.129 J/(g-°C), the specific heat of gold
Cw = 4.18 J/(g-°C), the specific heat of water
At equilibrium, heat lost by the gold - heat gained by the water.
Heat lost by the gold is
Qa = Ma*Ca*(T - Ta)
= (31.1 g)*(0.129 J/(g-°C)(*(69.3 - T °C)-
= 4.0119(69.3 - T) j
Heat gained by the water is
Qw = Mw*Cw*(T-Tw)
= (64.2 g)*(4.18 J/(g-°C))*(T - 27.8 °C)
= 268.356(T - 27.8)
Equate Qa and Qw.
268.356(T - 27.8) = 4.0119(69.3 - T)
272.3679T = 7738.32
T = 28.41 °C
Answer: 28.4 °C
Answer:
the correct answer is 17 significant figures
Answer: 1. 1.59 x 10^23 Particles
2. 4.79 x 10^21 particles
3. 2.67 x 10^25 particles
4. 2.12 x 10^23 Particles
Explanation: 1mole of any substance contains 6.02x10^23 Particles
1. 0.264 mol of silver will contain = 0.264 x 6.02x10^23 = 1.59 x 10^23 Particles
2. 7.95 x 10^-3 mol sodium chloride will contain = 7.95 x 10^-3 X 6.02x10^23 = 4.79 x 10^21 particles
3. 44.4 mol carbon dioxide will contain = 44.4 x 6.02x10^23 = 2.67 x 10^25 particles
4. 0.352 mol nitrogen will contain = 0.352 x 6.02x10^23 = 2.12 x 10^23 Particles
Answer: The weight of the object on earth =441N
Weight of the object on moon = 72N
