How much energy is required to heat 70 g of water at 20°C to boiling (100°C) if the specific heat of water is 4.184 J/g °C? Formula is

Question

How much energy is required to heat 70 g of water at 20°C to boiling
(100°C) if the specific heat of water is 4.184 J/g °C? Formula is Q=MCAT
29,288 J
23,430 J
5,858 J
35,146 J

in progress 0
Vân Khánh 3 days 2021-07-19T06:29:56+00:00 1 Answers 0 views 0

Answers ( )

    0
    2021-07-19T06:31:44+00:00

    Answer:

    Q=23,430J

    Explanation:

    Hello,

    In this case, since we compute the required energy via:

    Q=mC\Delta T

    Whereas m is the mass which here is 70 g, C the specific heat which for water is 4.184 J/(g°C) and ΔT is the temperature difference which is:

    \Delta T=100-20=80\°C

    Therefore, the energy turns out:

    Q=70g*4.184\frac{J}{g\°C}*80\°C\\ \\Q=23,430J

    Best regards.

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )