Question

Question #7c574

Answer 1

The specific heat capacity of your marshmallow is 0.33 cal·g⁻¹°C⁻¹

You use the formula `q = mcΔT`, but you first have to calculate the value of `q` and `ΔT`.

`q` = 30 s × `(100"cal")/(1"s")` = 3000 cal

`ΔT = T_2 – T_1` = (70 – 10) °C = 60 °C

`q = mcΔT`

`c = q/(mΔT) = (3000"cal")/"150 g × 60 °C"` = 0.33 cal·g⁻¹°C⁻¹