A 33.69 g sample of a substance is initially at 29.4 °c. after absorbing 1623 j of heat, the temperature of the substance is 110
.8 °c. what is the specific heat (sh) of the substance?
1 answer:
Q= mcΔT
1623 = 33.69g x c x (110.8 - 29.4)
1623 = 2742.366 g•°C x c
c = 0.59j/g•°C
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