50.0 grams of gold at a temperature of 85 degrees Celsius is added to a Styrofoam...

50.0 grams of gold at a temperature of 85 degrees Celsius is added to a Styrofoam cup calorimeter containing 150 grams of water. The cup of water is initially at a temperature of 25 degrees Celsius and the final equilibrium temperature of the water and gold mixture is 29 degrees Celsius. Calculate the specific heat of the gold. The specific heat of water is 4.18 J/g degrees Celsius. Please show the work too! I would greatly appreciate it!

Solutions

Expert Solution

Heat lose of gold = heat gain of water

mcT = mcT

50*c*(85-29) = 150*4.18*(29-25)

c = 0.9J/g-c⁰

The specific heat of gold = 0.9J/g-c⁰ >>>>>answer

queen_honey_blossom answered 1 year ago

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