In an experiment, 118.5 g of water is poured into a 18 g copper cup, both at an

Question

In an experiment, 118.5 g of water is poured into a 18 g copper cup, both at an initial temperature of 29.3 oC. Then a 75 g piece of steel at a temperature of 88.2 oC is immersed in the water. After a short time equilibrium is established without any heat loss to the surrounding air. (a) Calculate the final equilibrium temperature. oC Note the following constants: The specific heat for steel, copper, and water are 452 J/(kg*oC) , 387 J/(kg*oC) , and 4,186 J/(kg*oC) respectively. In an experiment, 118.5 g of water is poured into a 18 g copper cup, both at an initial temperature of 29.3 oC. Then a 75 g piece of steel at a temperature of 88.2 oC is immersed in the water. After a short time equilibrium is established without any heat loss to the surrounding air. (a) Calculate the final equilibrium temperature. oC Note the following constants: The specific heat for steel, copper, and water are 452 J/(kg*oC) , 387 J/(kg*oC) , and 4,186 J/(kg*oC) respectively. In an experiment, 118.5 g of water is poured into a 18 g copper cup, both at an initial temperature of 29.3 oC. Then a 75 g piece of steel at a temperature of 88.2 oC is immersed in the water. After a short time equilibrium is established without any heat loss to the surrounding air. (a) Calculate the final equilibrium temperature. oC Note the following constants: The specific heat for steel, copper, and water are 452 J/(kg*oC) , 387 J/(kg*oC) , and 4,186 J/(kg*oC) respectively.

Explanation / Answer

Heat energy gained or lost by an object = MC ( delta T) delta T is change in it's tempreture
Let equilibrium tempreture be T.
Heat energy gained by water = 118.5*4.186*(T-29.3)
Heat energy gained by copper = 18*0.387*(T-29.3)
Heat energy lost by steel = 75*0.452*(88.2 - T)
Total heat energy gained = total heat enrgy lost
118.5*4.186*(T-29.3) +18*0.387*(T-29.3) = 75*0.452*(88.2 - T)
503 *(T-29.3)= 33.9(88.2-T)
T = 33 oC

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