Question

A 58.87 g sample of a substance is initially at 20.5 °C. After absorbing 1169 J of heat, the temperature of the substance is 136.0 °C. What is the specific heat (c) of the substance?

Answer 1

To solve this problem, we need to consult the equation \[q(Heat)=m (Mass)*C (Specific Heat Capacity)* \Delta T\]. First, begin by stating your givens. Initial temperature = 20.5 degrees Celsius, and final temperature = 136.0 degrees Celsius. Heat absorbed (q) = 1169 J and the mass of the sample is 58.87 g. Next, we rearrange the equation to solve for our unknown, C (Specific Heat Cap.). q=m*C*(Tfinal-Tinitial) turns into \[C=\frac{ q }{ m*(Tfinal-Tinitial) }\] Finally, plug in our values to solve for the unknown heat capacity: C=(1169)/(58.87*(136.0-20.5)) = 0.1719... or 1.72*10^-1 with significant figures.