Question

an aluminum kettle weighs 1.05 kg how much heat and joules is required to increase the temperature of this kettle from 23° c to 99° c

Answer 1

Answer:

71820 J

Explanation:

From the question given above, the following data were obtained:

Mass (M) = 1.05 Kg

Initial Temperature (T₁) = 23 °C

Final temperature (T₂) = 99 °C

Heat (Q) required =?

Next, we shall determine the change in temperature. This can be obtained as follow:

Initial Temperature (T₁) = 23 °C

Final temperature (T₂) = 99 °C

Change in temperature (ΔT) =?

ΔT = T₂ – T₁

ΔT = 99 – 23

ΔT = 76 °C

Finally, we shall determine the heat required. This can be obtained as follow:

Mass (M) = 1.05 Kg

Change in temperature (ΔT) = 76 °C

Specific heat capacity (C) of aluminum = 900 J/KgºC

Heat (Q) required =?

Q = MCΔT

Q = 1.05 × 900 × 76

Q = 71820 J

Thus, 71820 J heat energy is required.