A 33.50 g sample of a substance is initially at 24.1 °C. After absorbing 1641 J of heat, the temperature of the substance is 117.6 °C. What is the specific heat (c) of the substance?
(1641 J) / (33.50 g) / (117.6 °C - 24.1 °C) = 0.5239 J/g·°C
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(1641 J) / (33.50 g) / (117.6 °C - 24.1 °C) = 0.5239 J/g·°C