Asu Physics Problems PDF
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Summary
These problems cover calculations related to heat transfer for various substances like water, ice, and steam. Sample calculations are provided for different scenarios, and the document demonstrates the application of formulas like Q = MCΔT.
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# Problem 1: - M<sub>cup</sub> = 500g - Heat of cup = 0.1 cal/g.C - t<sub>1</sub> = 20°C - t<sub>2</sub> = 96°C - M<sub>coffee</sub> = 200g - heat of water = 1 cal/g.C ## Answer: - Q<sub>m cup</sub> = M.C.Δt - 500 x 0.1 x (96 - 20) = 3800 cal - Q<sub>m coffee</sub> = M.C.Δt - 200 x 1 x (96 -...
# Problem 1: - M<sub>cup</sub> = 500g - Heat of cup = 0.1 cal/g.C - t<sub>1</sub> = 20°C - t<sub>2</sub> = 96°C - M<sub>coffee</sub> = 200g - heat of water = 1 cal/g.C ## Answer: - Q<sub>m cup</sub> = M.C.Δt - 500 x 0.1 x (96 - 20) = 3800 cal - Q<sub>m coffee</sub> = M.C.Δt - 200 x 1 x (96 - 20) = 15200 cal - Q<sub>m cup/coffee</sub> = 3800 + 15200 = 19000 cal # Problem 2: - M<sub>water</sub> = 180g - t<sub>1 water</sub> = 10 °C - t<sub>2</sub> = 90 °C - Equal temperature = 18 °C - Heat of water = 1 cal/g.C ## Answer: - Q<sub>m "water"</sub> = M.C.Δt - 80 x 1 x (18 - 10) = 11640 cal - Q<sub>m "cup"</sub> = M.C.Δt - 640 = M x 0.1 x (90 - 18) - M = 640 / (0.1 x (90 - 18)) = 88.89 g # Problem 3: - M "ice" = 10 g - t of steam = 100 °C - t of ice = 20 °C - Heat of warm water = 1cal/g.C - Heat (C) of Fusion = 80 cal/g.C - Heat of ice = 540 cal/g.C ## Answer: - Q<sub>m "ice"</sub> = M.C.Δt - 10 x 0.5 x 20 = 100 cal - Heat of Meadle ice = H<sub>ice</sub> x Δt + (F) - 10 x 80 = 800 cal - Heat to warm wodre From 0°C to 100°C - Q<sub>m "warm water"</sub> = M.C.Δt - 10 x 1 x 100 = 1000 cal - Heat to vaprice water - Q = H<sub>ice</sub> x Δt (v) = 10 x 540 = 5400 cal - Q '"totale"' = Q<sub>m "water"</sub> + Q<sub>mv</sub> + Q<sub>m "ica"</sub> + Q<sub>m "melting"</sub> - 1000 + 5400 + 100 + 800 = 6300 cal # Problem 4: - M<sub>stem</sub> = 49 g - t<sub>ice</sub> = 0 °C - t<sub>of steam</sub> = 100 °C - t<sub>of the mixture</sub> = 60 °C ## Answer: - Q<sub>steam</sub> = (M<sub>s</sub> x Δt) + (M<sub>ice</sub> x C<sub>uder</sub>) x Δt - 4 x 1 x (100 - 60) = 2160 + 160 = 2520 cal - Q<sub>ice</sub> = M<sub>ice</sub> x Δt<sub>F</sub> + M<sub>ice</sub> x (t<sub>F</sub> - 0) - M<sub>ice</sub> x (80 + 60) = 140 M<sub>ice</sub> - 2320 = 140 M<sub>ice</sub> - M<sub>ice</sub> = 2320 / 140 = 16.57 g
