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A student investigates a pure metal, X . The student takes a 100.0 g piece of metal X , heats it to 500.0°C , then places it on a 1000.0 g block of ice at 0.0°C . The ice partially melts, and the final temperature of the metal, ice, and melted water is 0.0°C . The student calculates the experimental value of the specific heat capacity of metal X and records it as 0.24 J/(g⋅°C) . Calculate the magnitude of the energy change (qmetal) of metal X during the experiment.

User Burtek
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1 Answer

27 votes
27 votes

Answer:

q_metal = -12000 J = -12 KJ

Here, the negative sign indicates that the energy is lost by the metal piece.

Step-by-step explanation:

The magnitude of energy change of the metal X can be given by the following formula:


q_(metal) = mC\Delta T

where,

m = mass of metal = 100 g

C = Specific Heat Capacity of metal X = 0.24 J/g.°C

ΔT = Change in Temperature of Metal Piece = 0° C - 500°C = -500°C

Therefore, using these values in the equation, we get:


q_(metal) = (100\ g)(0.24\ J/g^oC)(-500^oC)\\

q_metal = -12000 J = -12 KJ

Here, the negative sign indicates that the energy is lost by the metal piece.

User Jimkont
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