If a special type of ice has a specific heat capacity of 2060 J/kg K. How much heat must be absorbed by 2.6 kg of ice at -27.00 C to r

Question

If a special type of ice has a specific heat capacity of 2060 J/kg K. How
much heat must be absorbed by 2.6 kg of ice at -27.00 C to raise it up to
0.00 C, before any melting takes place?

A. 140,000 J
B. 290,000 J
C. 56,000 J
D. None of the above

in progress 0
Thanh Thu 3 years 2021-09-05T14:14:07+00:00 1 Answers 36 views 0

Answers ( )

  1. thucuc
    0
    2021-09-05T14:15:18+00:00
    Reply

    Answer:

    Q = 144612 Joules.

    Explanation:

    Given the following data;

    Mass = 2.6 kg

    Initial temperature = -27°C to Kelvin = 273 + (-27) = 246K

    Final temperature = 0°C to Kelvin = 273K

    Specific heat capacity = 2060 J/kgK.

    To find the quantity of heat absorbed;

    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.

    dt = T2 – T1

    dt = 273 – 246

    dt = 27 K

    Substituting the values into the equation, we have;

    Q = 2.6*2060*27

    Q = 144612 Joules.

Leave an answer

Browse

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