Question

A 30.60 g sample of a substance is initially at 25.6 °c. after absorbing 2811 j of heat, the temperature of the substance is 175.0 °c. what is the specific heat (c) of the substance?

Answer 1

okay its like whipping sweetheart it goes cold to hot from liquid to a solid substance so you would need a constant heat

Answer 2

Answer:

I got c = 0.615

Explanation:

To solve this you'll want to use the equation Q = mcΔT where ΔT is change in temperature, m is mass, Q is the heat in joules and c is the constant

So, 2811 = 30.60g x c x (175.0 - 25.6)

2811 / (30.60g x 149.4) = c

c = 0.615