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23 August, 15:46

4. How many calories of heat are required to heat 1370 g of water from 21.3°C to

89.5°C?

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Answers (1)
  1. 23 August, 16:03
    0
    The equation is:

    q = m. c. ΔT

    m = mass (1370 g)

    c = specific heat of water (4.18 J/gC)

    ΔT = temperature which T1 = 21.3C, T2 = 89.5C

    Applying all values to the equation above!

    q = 1370 g x 4.18 J/gC x (89.5 C - 21.3 C)

    :. q = 390554 J

    You need the answer in calories, so we should convert joules to calories by:

    J / 4.184

    So,

    390554 J / 4.184 = 93345 calories!
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