Question

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.

Answer 1

Answer:

q_metal = -12000 J = -12 KJ

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

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.