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Question 3, chap 120, sect 8. part 1 of 1 9 points A 0.0457 kg ingot...

Question 3, chap 120, sect 8. part 1 of 1 9 points A 0.0457 kg ingot of metal is heated to 238 ?C and then is dropped into a beaker containing 0.402 kg of water initially at 26?C. If the final equilibrium state of the mixed system is 28.4 ?C, find the specific heat of the metal. The specific heat of water is 4186 J/kg · ?C. Answer in units of J/kg · ?C.

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Answer #1

cm = specific heat of the metal = ?

cw = specific heat of water = 4186

T = final equilibrium temperature = 28.4

Tm = initial temperature of metal = 238

Tw = initial temperature of water = 26

m = mass of metal = 0.0457 kg

M = mass of water = 0.402 kg

using conservation of heat

energy lost by metal = energy gained by water

m cm (Tm - T) = M cw (T - Tw)

(0.0457) (cm) (238 - 28.4) = (0.402) (4186) (28.4 - 26)

cm = 421.62 J/kg-C

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