Question

360.0 g piece of iron is dropped into 425 mL of water at 24.00 °C. The final temperature of the water was measured as 45.70 °C. Calculate the initial temperature of the metal.

Answer 1

T(initial) = 241.4˚C

Step-by-step explanation:

∑q = q(metal) + q(water) = 0*

Q(metal) = (m∙c∙ΔT)Fe = 360g x 0.444j/g˚C x (45.7˚C – T(initial)) = 159.84j/˚C(45.7˚C – T(initial)) = 7304.69 j – 159.84j/˚C x T(initial)

Q(water) = (m∙c∙ΔT)water = 425g x 4.184j/g˚C x (45.7˚C - 24˚C) = 38,586.94 joules

∑[Q(metal) + Q(water)] = 7304.69j -159.84j/˚C x T(initial) + 38586.94j = 0

=> 159.84 x T(initial)j/˚C = 38586.94j

=> T(initial) = 38586.94j / 159.84j/˚C = 241.4˚C

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*Initial equation is from Heat Loss = Heat Gained (Conservation of Energy) That is, Heat lost + (- Heat gained) = 0

=> ∑(heat loss by metal) + (- heat gained by water) = 0

using heat transfer equation with temperature change

=> mcΔT(metal) + (-mcΔT(water) = mcΔT(metal) - (mcΔT(water) = 0