Question

A 13.5g sample of gold is heated, then placed in a calorimeter containing 60g of water. The temperature of water increases from 19oC to 20oC. The specific heat of gold is 0.13J/goC. What was the initial temperature of the gold metal sample?

Answer 1

Answer:

$T_i~=163.1$ ºC

Explanation:

We have to start with the variables of the problem:

Mass of water = 60 g

Mass of gold = 13.5 g

Initial temperature of water= 19 ºC

Final temperature of water= 20 ºC

Initial temperature of gold= Unknow

Final temperature of gold= 20 ºC

Specific heat of gold = 0.13J/gºC

Specific heat of water = 4.186 J/g°C

Now if we remember the heat equation:

$Q_H_2_O=m_H_2_O*Cp_H_2_O*deltaT$

$Q_A_u=m_A_u*Cp_A_u*deltaT$

We can relate these equations if we take into account that all heat of gold is transfer to the water, so:

$m_H_2_O*Cp_H_2_O*deltaT=~-~m_A_u*Cp_A_u*deltaT$

Now we can put the values into the equation:

$60~g*4.186~J/g{\circ}C*(20-19)~{\circ}C=-(13.5~g*0.13~J/g{\circ}C*(20-T_i)~{\circ}C)$

Now we can solve for the initial temperature of gold, so:

$T_i~=(\frac{60~g*4.186~J/g{\circ}C*(20-19)~{\circ}C}{13.5~g*0.13~J/g{\circ}C})+20$

$T_i~=163.1$ ºC

I hope it helps!

Answer 2

Taking into account the definition of calorimetry,  the initial temperature of the gold metal sample is 163.36 °C.

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, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case, you know:

• For gold:

1. Mass of gold = 13.5 g
2. Initial temperature of gold= Unknown
3. Final temperature of gold= 20 ºC
4. Specific heat of gold = 0.13 $\frac{J}{gC}$

• For water:

1. Mass of water = 60 g
2. Initial temperature of water= 19 ºC
3. Final temperature of water= 20 ºC
4. Specific heat of water = 4.186 $\frac{J}{gC}$

Replacing in the expression to calculate heat exchanges:

For gold: Qgold= 0.13 $\frac{J}{gC}$× 13.5 g× (20 C - Initial temperature of gold)

For water: Qwater= 4.186 $\frac{J}{gC}$× 60 g× (20 C - 19 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 gold gives up will be equal to the heat that the water receives. Therefore:

- Qgold = + Qwater

- 0.13 $\frac{J}{gC}$× 13.5 g× (20 C - Initial temperature of gold)= 4.186 $\frac{J}{gC}$× 60 g× (20 C - 19 C)

Solving:

- 1.755 $\frac{J}{C}$× (20 C - Initial temperature of gold)= 251.16 J

- 1.755 $\frac{J}{C}$× 20 C - (- 1.755 $\frac{J}{C}$)× Initial temperature of gold= 251.16 J

- 35.1 J +1.755 $\frac{J}{C}$× Initial temperature of gold= 251.16 J

1.755 $\frac{J}{C}$× Initial temperature of gold= 251.16 J + 35.1 J

1.755 $\frac{J}{C}$× Initial temperature of gold=286.7 J

Initial temperature of gold=286.7 J÷ 1.755 $\frac{J}{C}$

Initial temperature of gold= 163.36 °C

Finally, the initial temperature of the gold metal sample is 163.36 °C.

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