A 77.5 g sample of an unknown solid is heated to 62.5°C and placed into a calorimeter containing 93 g of water at 23.3°C. If the final temperature of the solid sample and the water is 26.2°C, what is the specific heat of the solid? Assuming no heat loss.

Specific heat of solid is 0.401J/g°C

Explanation:

In this experiment, the heat losed for the unknown solid is equal to heat gained for the water.

Using calorimeter equation:

Q = m×C×ΔT

Where Q is heat, m is mass of substance, C is specific heat and ΔT is change in temperature.

It is possible to write:

m[Solid] × C[Solid] × ΔT[Solid] = m[Water] × C[Water] × ΔT[Water]

Replacing:

77.5g × C[Solid] × (62.5°C - 26.2°C) = 93g × 4.184J/g°C × (26.2°C - 23.3°C)

77.5g × C[Solid] × (62.5°C - 26.2°C) = 93g × 4.184J/g°C × (26.2°C - 23.3°C)

2813.25g°C × C[Solid] = 1128.4J

C[Solid] = 0.401 J/g°C

Specific heat of solid is 0.401J/g°C

it is to be noted that an alpha particle is an helium nucleus. so it contains 2 protons 2 neutrons and has a mass number of 4. when alpha decay particles are ejected from the nucleus it gets transformed into an atom which has atomic number less than 2 and a mass number less than 4.

an equation that could represent alpha decay is radium 226 decay of alpha particle forms radon 222.

→

kdkdkrnmcnc

explanation:

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