Question

A 155.0 g piece of copper at 128 oC is dropped into 250.0 g of water at 17.9 oC. (The specific heat of copper is 0.385 J/goC.) Calculate the final temperature of the mixture. (Assume no heat loss to the surroundings.)

Answer 1

Step-by-step explanation:

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Answer 2

`T_(eq)=23.85^oC`

Step-by-step explanation:

Hello,

In this case, as the copper's heat loss is gained by the water, the following energetic relationship is:

`\Delta H_(Cu)=-\Delta H_(H_2O)`

Therefore the equilibrium temperature shows up as:

`m_(Cu)Cp_(Cu)(T_(Cu)-T{eq}) = m_(H_2O)Cp_(H_2O)(T_(eq)-T_(H_2O))\\\\T_(eq)=(m_(Cu)Cp_(Cu)T_(Cu)-m_(H_2O)Cp_(H_2O)T_(H_2O))/(m_(Cu)Cp_(Cu)-m_(H_2O)Cp_(H_2O)) \\`

Thus, by knowing that water's heat capacity is 4.18J/g°C, one obtains:

`T_(eq)=(155.0g*0.385(J)/(g^oC)*128^oC+250.0g*4.18(J)/(g^oC)*17.9^oC)/(155.0g*0.385(J)/(g^oC)+250.0g*4.18(J)/(g^oC))=23.85^oC`

Best regards.