Physics Past Paper PDF
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Uploaded by DelightedLesNabis702
جامعة الدلتا
Amal Elblasy
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Summary
This document contains physics problems and solutions related to heat transfer to various substances of different mass. Calculations and formulas are involved.
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# Page 1 ## Problem 1 - Cupper mug = 500g - Heat of capper = 0.1 cal/g.c - Coffee = 200g - Heat of water = 1 cal / g.c - Temperature = -20°C - Temperature = 96°C - Q<sub>mug</sub> = m<sub>1</sub>.C. (T<sub>f</sub> - T<sub>1</sub>) = 500 x 0.1 x (96 - 20) = 3800 cal - Q <sub>coffee</sub> = m<sub>...
# Page 1 ## Problem 1 - Cupper mug = 500g - Heat of capper = 0.1 cal/g.c - Coffee = 200g - Heat of water = 1 cal / g.c - Temperature = -20°C - Temperature = 96°C - Q<sub>mug</sub> = m<sub>1</sub>.C. (T<sub>f</sub> - T<sub>1</sub>) = 500 x 0.1 x (96 - 20) = 3800 cal - Q <sub>coffee</sub> = m<sub>2</sub>.C<sub>2</sub> (T<sub>f</sub> - T<sub>1</sub>) = 200 x 1 x (96 - 20) = 15200 cal - Q<sub>total</sub> = Q<sub>mug</sub> + Q<sub>coffee</sub> = 3800 + 15200 = 19000 cal ## Problem 2 - Mass of water = 80g - T of water = 10°C - T of caper = 90°C - Equilibrium temperature = 18°C - Heat of water = 1 cal/g.c - Q<sub>water</sub> = m<sub>w</sub>.C<sub>w</sub> x (T<sub>e</sub> - T<sub>water</sub>) = 80 x 1 x (18 -10) = 640 cal - Q<sub>capper</sub> = m<sub>c</sub> x C x (T<sub>i</sub> - T<sub>e</sub>) = m<sub>c</sub> x 0.1 x (90 - 18) = 7.2 m<sub>copper</sub> cal - 640 + 7.2 m<sub>copper</sub> = 640 = 88.89 g 7.2 # Page 2 ## Problem 3 - Mass of ice = 10g - T of steam = 100 °C - Temp of ice = 20°C - Heat of fusion = 80 cal/g - Heat of vaporization = 540 cal/g - Heat of warm from 20°C to 0°C - Q<sub>warming ice</sub> = m x c x (0 - (-20)) = 10 x 0.5 x 20 = 100 cal - Heat to melt ice - Q = m<sub>ice</sub> x L<sub>f</sub> = 10 x 80 = 800 cal - Heat to warm water from 0°C to 100°C - Q<sub>warming water</sub> = m<sub>ice</sub> x c<sub>water</sub> x (100 - 0) = 10 x 1 x 100 = 1000 cal - Heat to vaporize water - Q<sub>v</sub> = m<sub>ice</sub> x L = 10 x 540 = 5400 cal - Q<sub>total</sub> = Q<sub>water</sub> + Q<sub>v</sub> + Q<sub>ice</sub> + Q<sub>melting</sub> = 1000 + 540 + 100 + 800 = 7300 cal ## Problem 4 - Mass of steam = 4g - Temp of the mixture = 60°C - Temp of ice = 0°C - Temp of steam = 100 °C - Q<sub>steam</sub> = m<sub>s</sub> x L<sub>f</sub> + m<sub>ice</sub> x C<sub>water</sub> x (T<sub>steam</sub> - T<sub>f</sub>) = 4 x 1 x (100 - 60) = 2160 + 160 = 2320 cal - Q<sub>ice</sub> = m<sub>ice</sub> x L<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 = 140m<sub>ice</sub> - m<sub>ice</sub> = 2320 / 140 = 16.57g
