A237.0 g sample of molybdenum metal is heated to 100.10 °c and then dropped into an insulated cup containing 244.0 g of water at 10.00 °c. if the final temperature of the water and metal in the cup is 15.30 °c, then what is the specific heat of molybdenum? (specific heat of water = 4.186 j/g-°c do not add the unit in the answer.

Answer : The specific heat of molybdenum metal is,

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

where,

= specific heat of molybdenum metal = ?

= specific heat of water =

= mass of molybdenum metal = 237.0 g

= mass of water = 244.0 g

= final temperature of water and metal =

= initial temperature of molybdenum metal =

= initial temperature of water =

Now put all the given values in the above formula, we get

Therefore, the specific heat of molybdenum metal is,

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3

b) nuclear ive learned that in a pedtrodic table