Answers
a) Given, mass of metal = 8.6 g. Initial temp. of metal = 100 degree c Final temperature = 26.4 degree c Let, specific heat of the metal be c J/g degree c Heat lost by metal in water = 8.6 times c times (100 - 26.4)J. Initial temperature of water = 25 degree c Final temperature of water = 26.4 degree c mass of water = 402.4 g.
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Heat gained by water = 402.4 times 4.184 (26.4 - 25) Now according to law of conservation of energy, heat lost = heat gained => 402.4 times 4.184 times (26.4 - 25) = 8.6 times c times (100 - 26 - 4) => 2357.1 = 632.96 times c => c = 3.72 Therefore specific heat of the metal is 3.72 J/g degree c b) Initial temperature of the water-pot system = 24 degree c. Water will boil when temperature of both water and copper pot attains 100 degree c So, total heat energy required to boil =
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