A72.0-gram piece of metal at 96.0 °c is placed in 130.0 g of water in a calorimeter at 25.5 °c. the final temperature in the calorimeter is 31.0 °c. determine the specific heat of the metal. show your work by listing various steps, and explain how the law of conservation of energy applies to this situation.

The specific heat of the metal,  S = 0.1528 cal/g °C.

Explanation:

Given that, A 72.0-gram piece of metal at 96.0 °C is placed in 130.0 g of water in a calorimeter at 25.5 °C.

And the final temperature in the calorimeter is 31.0 °C.

By law of conservation of energy, energy is neither created nor destroyed.

So, energy lost by metal piece is equal to energy gained by water in calorimeter.

Specific heat of water is 1 cal/g °C

⇒ heat energy  Q = mSΔT , where m = mass of substance

                                                         S = specific heat

                                                         ΔT = change in temperature

Now, heat lost by metal piece, Q = 72×S×(96-31)

                                                       = 4680×S cal

Heat gained by water, Q = 130×1×(31-25.5)

                                         = 715 cal

⇒ 4680×S = 715.

⇒ S = 0.1528 cal/g °C.

the number of valence electrons for each molecule or ion is shown beneath the structure. remember hydrogen will not have more than two electrons. this hydrogen is part of a covalent bond (sharing two electrons). a double bond here would cause hydrogen to share four electrons with phosphorus.