Specific Heat A .500 kg sample of metal is heated to 345 C. The sample of metal
ID: 2012058 • Letter: S
Question
Specific Heat A .500 kg sample of metal is heated to 345 C. The sample of metal is placed into 890 kg of water that is 125 C. The final temperature of the sample and water is 234 C. Calculate the Cp of the sample and the Q of the water and sample. A French fry that has mass of 109 kg and has a temperature of 37 C is placed into a .210 kg cup of ketchup that has a temperature of 20 C. The final temperature of the fry and ketchup is 29 C. The heat lost by the French fry is 12 J. Calculate the Cp of the French fry. A silver bracelet with a mass of 350 kg is at body temperature (37C) and is subjected to an outside temperature of 10 C. The heat lost by the bracelet is 15 J. Calculate the Cp of the bracelet. A .540 kg sample of metal is heated to 345 C and placed into a 560 kg container of water that is cooled to 124 C. Calculate the amount of energy lost by the metal and gained by the water. Calculate the specific heat of the metal.Explanation / Answer
1. Given Mass of the metal , m1 = 0.5 kg Initial temperature of m1 is , Ti = 3450C Mass of the water , m2 = 0.890 kg Initial temperature of m2 is , Ti' = 1250C Final temperature of water and metal is, Tf = 2340C Specific heat of water, Cw = 4186 J/ kg 0C Applying conservation of energy Heat lost by metal = Heat gained by water m1 Cp (Ti - Tf) = m2 Cw (Tf - Ti) 0.5 kg * Cp* (3450C - 2340C) = 0.890 kg * 4186 J/ kg 0C* (2340C - 1250C) Cp = 7316.826 J/ kg 0C Thus, the specific heat of the sample is 7316.826 J/ kg 0C ----------------------------------------------------------------------------------------- ----------------------------------------------------------------------------------------- Heat gained by water , Q = m2 Cw (Tf - Ti) = 0.890 kg * 4186 J/ kg 0C* (2340C - 1250C) = 4.06 *10^5 J Heat lost by metal is, Q' = m1 Cp (Ti - Tf) = 0.5 kg *7316.826 J/ kg 0C * (3450C - 2340C) = 4.06 *10^5 J * post only question for single post *Related Questions
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