A copper calorimeter can with mass 0.100 kg contains 0.160 kgof water and 0.018 kg of ice in thermal equilibrium at atmospheric pressure. If 0.750 kg of lead at a temperature is dropped into the calorimeter can, what is thefinal temperature? Assume that no heat is lost to the surroundings.

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A copper calorimeter can with mass 0.100 kg contains 0.160 kgof water and 0.018 kg of ice in thermal equilibrium at atmospheric pressure.  If 0.750 kg of lead at a temperature  is dropped into the calorimeter can, what is thefinal temperature? Assume that no heat is lost to the surroundings.

Roosevelt Houghton · Accepted Answer

Mass of ice (mi)=0.018 kg  Latent Heat of ice ( Li ) = 333000 J / kg  Total heat for the melting of ice Qi=mi⋅Li  = 5994.0 J  Mass of Lead (mL) = 0.75 kg  Temp. of Lead (tL) = 255 C∘  Spec.heat capacity of Lead (cL)= 128 J / kg - K  Total heat comes from Lead QL=(mL)⋅(cL)⋅255  =24480 J  Since QL&amp;gt;Qi  So all ice will melt and the temperature will increase up to t∘ C  Mass of water (mw) = 0.16 kg  Temp. difference of water(△t)=t C∘  spec.heat capacity ofwater (Cw) = 4190 J / kg- K  Mass of copper (mcu)=0.100kg  Temp. difference of copper (△t)=t C∘  spec.heat capacity of copper (Ccu) = 386 J / kg-K  Apply:  Principle of calometry  miLi+(micw+mwcw+mcuccu)(t)=mLcL(225−t)  =5994.0J+(75.42+670.4+38.6)(t)=96(225−t)  =5994.0J+784.42t=21600−96t  21600−5994=(784.42+96)t  t=17.7∘ C

A copper calorimeter can with mass 0.100 kg contains 0.160 kgof water and 0.018 kg of ice in thermal equilibrium at atmospheric pressure. If 0.750 kg of lead at a temperature is dropped into the calorimeter can, what is thefinal temperature? Assume that no heat is lost to the surroundings.

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