a block of copper with mass 89.7g is heated to 148 °C and then dropped into 166mL water and allowed to come to equilibrium temperature. the water is initially 25 °C and contained in styrofoam cup . what is the final temperature?
specific heat capacity >copper=0.385J/g°C and water=4.184J/g°C
specific heat capacity >copper=0.385J/g°C and water=4.184J/g°C
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It's the old "conservation of energy" trick. The energy given off by the heated block of metal as it cools to thermal equilibrium is the same as the energy absorbed by the water as it warms to thermal equilibrium. Heat (the energy which is transferred) is given by q = mcΔT. ΔT = Tf - Ti. Let Tf = final equilibrium temperature. Th = temperature of the hot metal. Tc = temperature of the cold water.
-q(lost) = q(gained) ...... the negative sign corrects for the "direction" of the heat flow.
-mcΔT(metal) = mcΔT(water)
-mc(Tf - Th) = mc(Tf - Tc)
-89.7g x 0.385J/gC x (Tf - 148C) = 166g x 4.18J/gC x (Tf - 25C)
Tf = 30.8C
-q(lost) = q(gained) ...... the negative sign corrects for the "direction" of the heat flow.
-mcΔT(metal) = mcΔT(water)
-mc(Tf - Th) = mc(Tf - Tc)
-89.7g x 0.385J/gC x (Tf - 148C) = 166g x 4.18J/gC x (Tf - 25C)
Tf = 30.8C