Question

Calculate the specific heat capacity of a piece of ice if 1.30 kg of the wood absorbs 6.75×104 joules of heat, and its temperature changes from 32 ºC to 57 ºC.

Answer 1

Answer:

1.8  J/ g ∘ C

Explanation:

A substance's specific heat tells you how much heat much either be added or removed from  1 g  of that substance in order to cause a  1 ∘ C  change in temperature.

The equation that establishes a relationship between specific heat, heat added or removed.

Answer 2

Answer:

2076.9 J/kg°C

Explanation:

The equation needed here is:

$q=mC(T_f-T_i)$ , where q is the energy in Joules, m is the mass in kg, C is the heat capacity, T_f is the final temperature, and T_i is the initial temperature.

Here, we have q = 6.75*$10^4$ J, m = 1.30kg, T_f = 57, and T_i = 32. So:

$6.75*10^4=1.30*C*(57-32)$

Solving for C, we get:

C = 2076.9 J/kg°C

Hope this helps!