Question

4. If 13.34 g of an unknown substance has a -40.8 J of heat in cooling from 50°C to 43°C,

what is the specific heat of the unknown?

Answer 1

Answer: c (specific heat capacity) = 0.4369243949 J/g *C

Or with Significant digits: 0.44 ( 2 sig digs because of temperature)

Explanation: So this is a simple calorimetry question, so you have to consider this formula, ΔH = -Q and since Q = mcΔt || Therefore you get this equation ΔH = - (mcΔt). So you want to solve for c, now rearrange. Once you do that, you will end up with c = - (ΔH/ -m · -Δt)

Now Plug in the numbers

Then you will get 0.44!

Hope that helped and remember significant digits! Good luck!