35g of copper pellets are removed from a 300C oven and immediately dropped into

Question

35g of copper pellets are removed from a 300C oven and immediately dropped into 80ml of water at 24C in an insulated cup. What will the new water temperature be?

Explanation / Answer

Equation for heat: q = m · Cp · ?T where, q = heat m = mass Cp = specific heat ?T = (T higher – T lower) The trick for these type of problems, which involves substances being brought into contact, is to make all ?T positive; ?T = T higher - T lower. --------------------------------------… q lost from hot pellets = q gained in cool water --------------------------------------… q lost from hot pellets: q pellets = m · Cp · ?T where, m = mass Cu = 30.0 g Cp Cu = 385 J/kg·K = 385 J/kg·ºC x (1kg / 1,000g) = 0.385 J/kg·ºC Ti = 300.ºC ?T = (T higher – T lower) = (Ti – Tf) = (300.ºC – Tf) Substituting numbers into the equation: q pellets = (30.0 g) · (0.385 J/g·ºC) · (300.ºC – Tf) ==> q pellets = 3,465 J – 11.50 J/ºC · Tf ~(eq1) --------------------------------------… q gained in cool water: q water = m · Cp · ?T where, assuming density of water = 1g/mL m water = 100.mL x (1g/mL) = 100.g Cp water = 4190 J/kg·K = 4190 J/kg·ºC x (1kg / 1,000g) = 4.190 J/g·ºC Ti = 20.0ºC ?T = (T higher – T lower) = (Tf – Ti) = (Tf – 20.0ºC) Substituting numbers into the equation: q water = (100.g) · (4.190 J/g·ºC) · (Tf – 20.0ºC) ==> q water = 419 J/ºC · Tf – 8,380 ~(eq2) --------------------------------------… q lost from hot pellets = q gained in cool water ==> ~(eq1) = ~(eq2) 3,465 J – 11.50 J/ºC · Tf = 419 J/ºC · Tf – 8,380 J Solving for Tf: 430.5 J/ºC · Tf = 11,845 J Tf = 11,845 J / 430.5 J/ºC = 27.515 ºC *** Answer: 27.5 ºC ***

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