Question

PHYS 205 Winter 2019 Lab 2 Homework 5. A student mixed 100 g of liquid A at a temperature of 80 °C with 100 g of liquid B at a temperature of 20 °C. After the mixture comes to thermal equilibrium it has a temperature of 40 °C. The student contends: "If I mix 100 g of liquid A at 60 °C with 100 g of liquid B at 30 °C, I will also end up with a 200 g mixture at 40 °C What, if anything, is wrong with the student's contention? If something is wrong, identify it and explain how to correct it. If nothing is wrong, explain the physics behind the students answer. How much energy does it take to melt 10 g of ice and then warm the resulting water to 20 °C? (show your work) 6. 7. When you added salt to your ice water slush mix the temperature went down, why? If the temperature goes down, how is it possible that sprinkling salt on icy sidewalks helps to melt the ice? (Use plain language to explain this from an energy perspective)

Answer 1

Log of ice longrightarrow log of water at 20 degree c log of ice longrightarrow 10 of water at 0 degree c longrightarrow g of water at 20 degree h_1 = heat of fusion = 335 J/gram = 3350 J/10 gram therefore for log of ice, h_1 = 3350 J h_2 = specific heat = 4180 J/ks = 4.180 J/g degree c h_2 = 41.80 times 20 J/(log) (20 degree c) h_2 = 836 J Total energy = h_1 + h_2 = 4186 J Energy lost by hot water = energy by cold water m_A c_A (T_not - x) = m_B c_B (x - T_cold) rightdoublearrow 100 c_A (80- 40) = 100 c_B (40 - 20) c_A (10)^2 = c_B (20) 100 c_a (60 degree - 40 degree - 100 c_B (40 - 30) c_A 20 = c_B = 20 c_A/c_B = 1/2 Here Liquid A has different specific heats from liquid B 7) is so cold that it freezed the ice (cream) mixture dissolving longrightarrow used to melt the ice.