A30.5 g sample of an alloy at 95.0°c is placed into 49.3 g water at 24.3°c in an insulated coffee cup. the heat capacity of the coffee cup (without the water) is 9.2 j/k. if the final temperature of the system is 31.1°c, what is the specific heat capacity of the alloy? (c of water is 4.184 j/g×k)

0.752 J/g*K

Explanation:

The heat lost by the alloy (which is negative) must be equal to the heat gained by the water and the coffee cup:

-Qa = Qw + Qc

-ma*ca*ΔTa = mw*cw*ΔTw + C*ΔTc

Where, m is the mass, c is the specific heat capacity, C is the heat capacity of the coffee cup, ΔT is the change in temperature, a represents the alloy, and w the water.

The coffee cup has initial temperature equal to the water, then:

-30.5*ca*(31.1 - 95.0) = 49.3*4.184*(31.1 - 24.3) + 9.2*(31.1 - 24.3)

1948.95ca = 1465.20

ca = 0.752 J/g*K

heat or electricity in the atmosphere- c.

answer: the rate of diffusion increases with increase in temperature.

increase in temperature, increases the molecular/particles speed (kinetic energy. diffusion means movement of particles through the material.

generally, the heavier the particle the more it interacts with the environment and more the temperature matters. so higher temperature results in higher diffusion.