A block of aluminum weighing 100g is cooled from 108.4°C to 68.2°C with the release of 1080 joules of heat. From this data, calculate the
specific heat of aluminum

Specific heat of aluminum  = 0.27 j/g.°C

Explanation:

Given data:

Mass of Al = 100 g

Initial temperature = 108.4°C

Final temperature = 68.2°C

Heat released = -1080 j

Specific heat of aluminum = ?

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 =  68.2°C -  108.4°C

ΔT = - 40.2°C

Q = m.c. ΔT

-1080 j = 100 g ×c ×- 40.2°C

-1080 j = -4020 g.°C ×c

c = -1080 j/-4020 g.°C

c = 0.27 j/g.°C

beta decay is the answer

1. a)c-14 has two more protons

2. d)they have different numbers of protons, neutrons, and valence electrons; they are all gases at room temperature.

3. c) iron and nickel.