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

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

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Adela 1 year 2021-08-30T14:17:38+00:00 1 Answers 17 views 0

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    2021-08-30T14:18:55+00:00

    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.

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