What is the final temperature of the copper and water given that the specific heat of copper is 0.385 #J/(g * "^oC)#?

Question

A hot lump of 48.7 g of copper at an initial temperature of 76.8 C is placed in 50.0 mL of #H_2O# initially at 25.0°C and allowed to reach thermal equilibrium. Assume no heat is lost to surroundings.

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Answer 1 · 2017-07-15T23:26:22

#T_f = 29.3 °C#

Here, we're setting the heat values derived given the equation:

#q = mC_sDeltaT#

equal to each other, as such,

#q_(water)+q(Cu) = 0#
#therefore q_(water) = -q(Cu)#

Before we start, I will assume the density of water is #"1.00 g/mL"#.

#50.0 g*(4.184J)/(g*°C)*(T_f-25.0°C) = -[48.7g*(0.385J)/(g*°C)*(T_f-76.8°C)]#

#209.2T_f - 5230 °C = -18.75T_f + 1440 °C#

#227.95T_f = 6670 °C#

#therefore T_f = 29.3°C#