Answer:
The specific heat capacity for gold in 105 joules which are required to heat 30.0 grams of gold is 0.129 J/(g℃)
Explanation:
We make use of the formula
[tex]Q=m \times c \times \Delta T[/tex]
where
∆T = final T - initial T
= 54.9℃ - 27.7℃ = 27.2℃
Q is the heat energy in Joules = 105J
c is the specific heat capacity = ?
m is the mass of Gold = 30.0g
[tex]Q=m \times c \times \Delta T[/tex]
Rearranging the formula
[tex]c= \frac {Q}{(m\times \Delta T)}[/tex]
[tex]= \frac {105J}{(30.0g \times 27.2 ^\circ{C})}\\\\= \frac {105J}{(816g^\circ{C})}[/tex]
So,
c = 0.129 J/(g℃)
(Answer)
