Question #a5e2b

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Question #a5e2b

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Answer 1 · 2017-03-29T05:56:09

$\approx 0.127 \frac{J}{g \cdot K}$ $~~0.127J/(g*K)$

Explanation:

$Heat absorbed = mass of substance \cdot specific heat capacity of substance \cdot δ T$ $"Heat absorbed"="mass of substance"*"specific heat capacity of substance"*delta T$

$1217 J = 55.86 g \cdot specific heat of substance \cdot ([273 + 196] - [273 + 24])$ $1217J=55.86g*"specific heat of substance"*([273+196]-[273+24])$

$Specific heat of substance = \frac{1217}{55.86 \cdot ([273 + 196] - [273 + 24])} \approx 0.127 \frac{J}{g \cdot K}$ $"Specific heat of substance"=1217/(55.86*([273+196]-[273+24]))~~0.127J/(g*K)$

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