Ten grams of iron cooled from 60.4 °C to 35.0°C, and released 114 J of heat in the process. What is the specific heat of iron?
Your tool of choice here will be the equation
#color(blue)(|bar(ul(color(white)(a/a)q = m * c * DeltaTcolor(white)(a/a)|)))" "#, where
Now, because your sample of iron is being cooled, the heat it released will actually carry a negative sign.
#q = -"114 J"#
That happens because the final temperature of the sample is lower than the initial temperature, which makes the change in temperature negative.
You will have
#DeltaT = 35.0^@"C" - 60.4^@"C" = -25.4^@"C"#
All you have to do now is rearrange the equation to solve for
#q = m * c * DeltaT implies c = q/(m * DeltaT)#
Plug in your values to find
#c = (-"114 J")/("10 g" * (-25.4^@"C")) = color(green)(|bar(ul(color(white)(a/a)color(black)("0.45 J g"^(-1)""^@"C"^(-1))color(white)(a/a)|)))#
I'll leave the answer rounded to two sig figs.
Iron's specific heat is listed as being equal to