Question

How much heat is released as a 5.89 kg block of aluminum cools from 462 °C to 315 °C. The specific heat capacity of aluminum is 0.900 J/(g*K).

Answer 1

779,247 J is the amount of heat released.

Explanation:

The equation that gives the amount of heat supplied is

                 $E \text { or } Q=m \times c_{A l} \times \Delta T$

Where,

E or Q – amount of heat supplied

m- mass

$c_{A l}$ - specific heat capacity of Aluminium

$\Delta T$ – temperature variation caused by heat change

The amount of heat energies that causes the temperature to vary $1^{\circ} \mathrm{C}$ or 1 K per kg of the material is known as specific heat capacity (c)

Here, Given data:

$c_{A l}$  – 0.900 J/g.K

To convert gram into kilogram, multiply and divide by $10^{3}$, we get  

Specific heat capacity of aluminum = 900 J/kg. K

m – 5.89 kg

$\Delta T=462^{\circ}-315^{\circ}=147^{\circ} \mathrm{C}$

By substituting the known values in the above equation, we get

$E \text { or } Q=5.89 \times 900 \times 147=779,247 \mathrm{J}$

Answer 2

Answer: C) -779,000

Explanation: First convert the mass to grams.

m=(5.89 kg)*(1000 g/1 kg)= 5890 g

Then find the change in temperature. It is not necessary to convert the temperature to Kelvin because the change in temperature is the same in both Kelvin and Celsius.

ΔT= (Final - Initial)= 315 °C - 462 °C= -147 °C

Now put these into the heat equation.

q=mcΔT= (5890 g)*(0.900 J/(g*K))*(-147 °C)= -779247 J= -779000 J