Apiece of unknown metal with mass 68.6 g is heated to an initial temperature of 100 °c and dropped into 42 g of water (with an initial temperature of 20 °c) in a calorimeter. the final temperature of the system is 52.1°c. the specific heat of water is 4.184 j/g*⁰c. what is the specific heat of the metal?

The piece of unknown metal is in thermal equilibrium with water such that Q of metal is equal to Q of the water. We write this equality as follows:

-Qm = Qw

Mass of metal (Cm)(ΔT) = Mass of water (Cw) (ΔT)

where C is the specific heat capacities of the materials.

We calculate as follows:

-(Mass of metal (Cm)(ΔT)) = Mass of water (Cw) (ΔT)

-68.6 (Cm)(52.1 - 100) = 42 (4.184) (52.1 - 20)

Cm = 1.717 > OPTION C

the cold front replaces the warmer air at the ground level changing the wind to be breezy and denser.there is also an immediate drop in the temperature.