A metal object with mass of 20.4 g is heated to 97.0 °C and then transferred to an insulated container containing 80.9 g of water at 20.5 °C. The water temperature rises and the temperature of the metal object falls until they both reach the same final temperature of 22.7 °C. What is the specific heat of this metal object? Assume that all the heat lost by the metal object is absorbed by the water.

Explanation:

Heat lost by metal object = mass x specific heat x fall in temperature

= 20.4 x S x ( 97 - 22.7 )

= 1515.72 S

Heat gained by water

= 80.9 x 1 x ( 22.7 - 20.5 )

= 177.98 Cals

heat lost = heat gained

1515.72 S = 177.98

S = .117 Cals /g . C

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂

layers of the earth affect the waves and these effects are seen on instruiments such as seimometers