A copper pot with a mass of 0.500 kg contains 0.170 kg of water,and both are at a temperature of 20.0 ∘C . A 0.250 kg block of ironat 85.0 ∘C is dropped right into the pot.
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Find the last temperature of the system, assuming no warmth lossto the surroundings.
The idea used to solve this difficulty is the heat equation in thermodynamics.
Initially, use the idea of equilibrium temperature of the mechanism. Then, uncover the final temperature of the system by equating the warm shed by the iron block through the warmth got by the copper and also the water. Finally, rearvariety the expression for the temperature and then calculate.
From the zeroth law of thermodynamics when 2 objects are areas in contact warmth is move from high temperature to low temperature until they reach the exact same temperature.
The iron block has actually better temperature than copper pot contain water. Hence the temperature will carry from iron block to copper pot and also water.
The expression for the equilibrium temperature of the mechanism is,
The expression for the amount of warm forced to readjust the temperature is,
The expression for the amount of warm required to change the temperature is,
Expression for the amount of warmth lost by the iron block is,
The heat lost is,
Expression for the amount of heatobtained by water is,
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The warmth got by the water is,
The expression for the amount of heat gained by copper is,