heat problems to post.pptx - Heat of a chemical reaction...

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Final Temperature of water and object = 45.6 ºC Initial Temperature of water = 23.0 ºC Initial Temperature of 100. g object = 65.2 ºC Heat of a chemical reaction An unknown object weighing 100. g is heated to 65.2ºC. It is dropped into 250. mL of water at 23.0ºC. Thermal equilibrium is achieved at 45.6ºC. What is the specific heat capacity of the object? Assume the beaker did not absorb any heat. 250. mL of water
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Important concept: heat given off from the object is transferred to the water (q object =-q water ) 250. mL = 250. g WHY? Temperature of the water went up, therefore the heat taken in by the water is positive. Temperature of the object went down (it gave off heat), therefore the heat of the object is negative. ∆T=T final - T initial q object = m object c s, object ∆T object q object =-2.36x 10 4 J (see statement above) Specific heat capacity MUST be positive q water = ∆T water m water c s, water q water = (45.6°C-23.0°C)(250. g)(4.184 ) q water = 2.36x10 4 J gained by the water An unknown object weighing 100. g is heated to 65.2ºC. It is dropped into 250. mL of water at 23.0ºC. Thermal equilibrium is achieved at 45.6ºC.
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