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!