Question

A 6.23 g nugget of pure gold absorbed 282 J of heat. The initial temperature was 30.4 degrees Celsius. What was the correct final temperature of the gold nugget if the specific heat of gold is 0.129 J/gC

Answer 1

Final temperature, T2 = 381.28°C

Step-by-step explanation:

Given the following data;

Mass = 6.23 g

Initial temperature = 30.4°C

Heat capacity = 282 J

Specific heat capacity = 0.129 J/g°C

To find the final temperature;

Heat capacity is given by the formula;

`Q = mcdt`

Where;

Q represents the heat capacity or quantity of heat.

m represents the mass of an object.

c represents the specific heat capacity of water.

dt represents the change in temperature.

Making dt the subject of formula, we have;

`dt = \frac {Q}{mc}`

Substituting into the equation, we have;

`dt = \frac {282}{6.23*0.129}`

`dt = \frac {282}{0.8037}`

dt = 350.88°C

Now, the final temperature T2 is;

But, dt = T2 - T1

T2 = dt + T1

T2 = 350.88 + 30.4

Final temperature, T2 = 381.28°C