You placed 43.1 g of an unknown metal at 100 °c into a coffee cup calorimeter that contained 50.0 g of water that was initially at 22.0 °c. the equilibrium temperature of mixing (t0) was determined to be 23.2 °c. the calorimeter constant was known to be 51.5 j/°c. specific heath2o = 4.184 j/g·°ca. what is the total amount of heat (j) lost by the metal? ng 1.5b. what was the specific heat (j/g·°c) of the metal? ng 1.5

Answer :

(a) The heat released by the metal is -312.48 J

(b) The specific heat of the metal is

Explanation :

For part A :

Heat released by the metal = Heat absorbed by the calorimeter + Heat absorbed by the water

where,

q = heat released by the metal

= heat absorbed by the calorimeter

= heat absorbed by the water

= specific heat of calorimeter =

= specific heat of water =

= mass of water = 50.0 g

= change in temperature =

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

Thus, the heat released by the metal is -312.48 J

For part B :

q = heat released by the metal = -312.48 J

m = mass of metal = 43.1 g

c = specific heat of metal = ?

= change in temperature =

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

Thus, the specific heat of the metal is

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