A 160-g metal container, insulated on the outside, holds 80 g of water in therma

Question

A 160-g metal container, insulated on the outside, holds 80 g of water in thermal equilibrium at 22.00°C. A 15-g ice cube, at the melting point, is dropped into the water, and when thermal equilibrium is reached the temperature is 15.00°C. Assume there is no heat exchange with the surroundings. For water, the specific heat is 4190 J/kg·K and the heat of fusion is 3.34 × 105 J/kg. The specific heat for the metal is closest to:

A. 4820 J/Kj*k

B. 3220 J/Kj*k

C. 3820 J/Kj*k

D. 2220 J/Kj*k

E. 4320 J/Kj*k

Please show all work and specify which answer choice is the correct one. Please do not present an answer that is not one of the choices.Thank you!!!!

Explanation / Answer

Heat gain by ice = m*Lf + m*C*delta T = 15*334 + 15*4.190*15 = 5010 + 942.75 = 5952.75

Heat lost by metal and water = mass of metal*specific heat + mass of metal * specific heat of water)* change in temperature

5952.75 = (160*C + 80*4.190)*7

5952.75 / 7 = 160*C + 335.2

850.39 = 160*C + 335.2

160*C = 515.19

C = 3.22

specific heat of metal = 3220 J/kg

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