Thermal Equilibrium and Heat Flow

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Presentation transcript:

Thermal Equilibrium and Heat Flow Lesson 3

Q heat lost = Q heat gained Thermal Equilibrium Q heat lost = Q heat gained

Q lost by steel = Q gain by water Example 5 20 g of steel with the temperature of 90°C is dropped into 0.25 kg of water with the temperature of 24 °C. What is the temperature when the steel and water reach thermal equilibrium? Specific heat capacity of water = 4,200 J/kg °C Specific heat capacity of steel = 450 J/Kg °C Solution: mstell = 0.02 kg cstell – 450 J/kg °C mwater = 0.25 kg cwater = 4,200 J/kg °C Q lost by steel = Q gain by water 0.02 x 450 x (90 –y ) = 0.25 x 4,200 x (y – 24) 9 x (90-y) = 1050 x (y-24) 810-9y = 1050 y – 25,200 25,200 + 810 = 1050 y + 9y 26,010 = 1059 y y = 26,010/1059 y = 24.56 °C

Heat Flow

Benefits of Heat Flow

Test Yourself Answer Textbook page 152

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