I have a few question please help its due in a few hours 1. Determine the specific heat of a certain metal if a 450 gram sample of it loses 34,500J of heat as its temperature drops by 97K 2. The temperature of a 25 gram sample of a certain metal drops by 103 K as it loses 2600 joules of heat. What is the specific heat of the metal?

3.What is the final temperature after 840 Joules is absorbed by 10.0g of water at 298K?
(please show steps)

c = 0.79 j/g.K

c = 1 j/g.K

T2 = 123.1 K

Explanation:

1)

Given data:

Mass of metal = 450 g

Heat loses = 34,500 j

Temperature drop = ΔT  = 97 K

Specific heat of metal = ?

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:

Mass of metal = 25 g

Heat loses = 2600 j

Temperature drop = ΔT  = 103 K

Specific heat of metal = ?

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

answer

i think the answer is a) as air pressure in an area increases, the density of the gas particles in that area decreases.

hope it