Question

An unknown solid object with a mass of 35.0 g was heated to 100.0 °C and transferred to a calorimeter containing 50.0 g of water at 23.0 °C. After thermal equilibrium was reached, the temperature inside the calorimeter was 26.5 °C. Calculate the specific heat of the unknown solid (specific heat of water = 4.184 J/g°C).

Answer 1

Taking into account the definition of calorimetry, the specific heat of the unknown solid is 0.285 $\frac{J}{gC}$.

What is calorimetry

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

So, the equation that allows to calculate heat exchanges is:

Q = c× m× ΔT

where:

• Q is the heat exchanged by a body of mass m
• c is the specific heat of the substance c
• ΔT=Final temperature - Initial temperature is the temperature variation.

Specific heat of the unknown solid

In this case, you know:

• For the unknown solid:

1. Mass of solid = 35 g
2. Initial temperature of solid= 100 °C
3. Final temperature of solid= 26.5 ºC
4. Specific heat of solid= unknown

• For water:

1. Mass of water = 50 g
2. Initial temperature of water= 23 ºC
3. Final temperature of water= 26.5 ºC
4. Specific heat of water = 4.186 $\frac{J}{gC}$

Replacing in the expression to calculate heat exchanges:

For unknown solid: Qsolid= csolid × 35 g× (26.5 - 100)°C

For water: Qwater= 4.186 $\frac{J}{gC}$× 50 g× (26.5 - 23)°C

If two isolated bodies or systems exchange energy in the form of heat, the quantity received by one of them is equal to the quantity transferred by the other body. That is, the total energy exchanged remains constant, it is conserved.

Then, the heat that the solid gives up will be equal to the heat that the water receives. Therefore:

- Qsolid = + Qwater

Replacing the corresponding expressions:

- csolid × 35 g× (26.5 - 100)°C= 4.186 $\frac{J}{gC}$× 50 g× (26.5 - 23)°C

Solving:

csolid × 2,572.5 g×°C= 732.55 J

csolid =732.55 J÷ 2,572.5 g×°C

csolid=0.285 $\frac{J}{gC}$

Finally, the specific heat of the unknown solid is 0.285 $\frac{J}{gC}$.

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