Acalorimeter contains 124 g of water at 26.6°c. a block of metal with a mass of 26 g is heated to 95.8°c and then placed in the water in the calorimeter. after sufficient time, the temperature of the water is measured and found to be 29.0°c. calculate the heat capacity per gram of metal. assume no heat is lost to the calorimeter or the surroundings.

The heat capacity of the metal is 0.717 J/g°C

Explanation:

Step 1: Data given

Mass of water = 124.0 grams

Mass of metal = 26.0 grams

Temperature of water = 26.6 °C

Temperature of metal = 95.8 °C

Final temperature of the water = 29.0 °C

Step 2: Calculate heat capacity of metal

Qlost = -Q gained

Qmetal = -Qwater

Q = m*c*ΔT

⇒ with m = mass in grams

⇒ c= the heat capacity in J/g°C

⇒ ΔT = The change in temperature = T2 - T1 (in °C)

Qmetal = -Qwater

m(metal)*c(metal)*ΔT(metal) = -m(water)*c(water)*ΔT(water)

26 *c(metal) *(29-95.8) = -124 * 4.184 *(29-26.6)

-1736.8 * c(metal) = -1245.16

c(metal) = 0.717 J/g°C

The heat capacity of the metal is 0.717 J/g°C

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.