Question

If 63.4 J of heat are added to a sample
of aluminum with a mass of 16.33 g, what is
its temperature change? The specific heat of
aluminum is 0.899 J/g °C

Answer 1

Answer:

dt = 4.32°C

Explanation:

Given the following data;

Mass = 16.33g

Quantity of heat, Q = 63.4J

Specific heat capacity = 0.899 J/g °C

To find the temperature change;

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;

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

Substituting into the equation, we have;

$dt = \frac {63.4}{16.33*0.899}$

$dt = \frac {63.4}{14.6807}$

dt = 4.32°C

Therefore, the change in temperature for this aluminum is equal to 4.32°C.