What is the specific heat ( π ) of the substance?
2021-04-02 1:06 pm
A 46.41 g sample of a substance is initially at 27.5 Β°C. After absorbing 1837 J of heat, the temperature of the substance is 130.1 Β°C. What is the specific heat ( π ) of the substance?
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2021-04-02 2:06 pm
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Heat absorbed = m c ΞT1837 J = (46.41 g) Γ c Γ [(130.1 - 27.5) Β°C]
c = 1837 / (46.41 Γ 102.6) J/gΒ°C
Specific heat, c = 0.3858 J/gΒ°C
ζΆιζ₯ζ: 2021-04-24 09:51:37
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