Hi, I need help with the following problem. Thanks! An aluminum cube of mass 44.

Question

Hi, I need help with the following problem. Thanks! An aluminum cube of mass 44.25g is immersed in liquid Nitrogenand comes to thermal equilibrium with the liquid Nitrogen. Thisequilibrium temperature is that of liquid Nitrogen,-198.5oC. The cube is then immersed in 189.47g of waterat 22.3oC. The final equilibrium temperature of thewater-aluminum cube system is 14.0oC. Calculate thespecific heat of the aluminum cube. cAl = ___________________ (Answer should have 2 numbers afterthe decimal point.) Hi, I need help with the following problem. Thanks! An aluminum cube of mass 44.25g is immersed in liquid Nitrogenand comes to thermal equilibrium with the liquid Nitrogen. Thisequilibrium temperature is that of liquid Nitrogen,-198.5oC. The cube is then immersed in 189.47g of waterat 22.3oC. The final equilibrium temperature of thewater-aluminum cube system is 14.0oC. Calculate thespecific heat of the aluminum cube. cAl = ___________________ (Answer should have 2 numbers afterthe decimal point.)

Explanation / Answer

            wehave the condition of method of mixtures, the heat lost by the coldsubstance is equal to the heat gain by the cool substance when twosubstances are mixed.             Thereforehere heat lost by water is equal to the heat gain by aluminiumcubes.    But heat lost or gain is given by Q =mct    Therefore we should havemAlcAl(14oC-(-198.5oC))=mWcW(22.3oC-14oC)             or(44.25g)cAl(212.5oC) =(189.47g)(4.186J/g/oC)(8.3oC)             solvingthe equation we get cAl = 0.7J/g/oC

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