Answer: The correct answer is 6.8 g.
Explanation
Step 1
The first step is to realize that when the ring is put in water, it will loose heat to the water. This process will decrease the temperature of the ring and increase the temperature of the water until the point where the ring and the water reach thermal equilibrium at 38 C.
Step 2
The next step is to calculate the amount of heat given out by the ring to heat up the water. The equation for the amount of heat that has to be lost to change the temperature of an object is [tex]Q=mc\Delta T[/tex] where [tex]Q[/tex] is the heat transferred, [tex]m[/tex] is the mass, [tex]c[/tex] is the specific heat capacity and [tex]\Delta T[/tex] is the change in temperature. The specific heat capacity of gold is 0.129J/gC, the mass of the ring is 47g and the rings temperature drops from 99 C to 38 C. Using this data, we calculate the heat transferred to the water from the ring as shown below,
[tex]Q=mc\Delta T\\Q=47g\times (0.129J/gC)\times(99C-38C)\\Q=369.84J[/tex]
Step 3
The next step is to take the equation for heat transferred and solve it for the mass. The specific heat capacity of water is 4.186J/gC, the temperature of the water increases from 25 C to 38 C. The amount of heat transferred to the water is 369.84 J, i.e the heat given out by the ring. The mass of water is calculated as shown below,
[tex]Q=mc\Delta T\\\\\implies m=\frac{Q}{m\Delta T} \\\\\implies m=\frac{369.84J}{(4.186J/gC)\times(38C-25C)} =6.8g[/tex]
The mass of water needed is 6.8g
