10.) 69% of all copper atoms are Cu-63. How many times more massive than an atom of C-12 is an atom this copper isotope?

1 = c = 0.79 j/g.K

2= c = 1 j/g.K

3 = T2 = 123.1 K

Explanation:

Given data:

Specific heat of metal = ?

Mass of metal = 450 g

Heat loses = 34,500 j

Temperature drop = ΔT  = 97 K

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

34500 j = 450 g ×c × 97 K

34500 j = 43650 g.K ×c

c = 34500 j/43650 g.K

c = 0.79 j/g.K

2)

Given data:

Specific heat of metal = ?

Mass of metal = 25 g

Heat loses = 2600 j

Temperature drop = ΔT  = 103 K

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

2600 j = 25 g ×c × 103 K

2600  j = 2575 g.K ×c

c = 2600 j/2575 g.K

c = 1 j/g.K

3)

Given data:

Mass of water = 10 g

Heat absorbed  = 840 j

Initial temperature drop = 298 K

Final temperature = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2 - T1

840 j = 10 g ×4.18 j/g.K × [T2 - 103 K]

840 j = 41.8 J/K ×  [T2 - 103 K]

840 J/41.8 J/ K = T2 - 103 K

20.1 k = T2 - 103 K

T2 = 20.1 K + 103 K

T2 = 123.1 K