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.

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:

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 = -12000 J = -12 KJ

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

a  solution  is always transparent, light passes through with no scattering from solute particles which are molecule in size. the  solution  is homogeneous and does not settle out. a colloid is intermediate between asolution  and a  suspension. while asuspension  will separate out a colloid will not.

the answer for sodium is 0.186

the answer for rubidium is 0.251