1. what is the specific heat of a substance that absorbs 2500 joules of heat when a sample of 100
g of the substance increases in temperature from 10 °c to 70°c?

0.4167 J/°C.g.

Explanation:

Q = m.c.ΔT,

where, Q is the amount of heat absorbed by substance (Q = 2500 J).

m is the mass of the substance (m = 100 g).

c is the specific heat of the substance (c = ??? J/°C.g).

ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = 70°C - 10°C = 60°C).

∵ Q = m.c.ΔT

∴ c = Q/(m.ΔT) = (2500 J)/(100 g)(60°C) = 0.4167 J/°C.g.

it’s previous models   because when i picked that son of a gun’s answer it was wrong look for previous models because they might make it a different letter  

12 is it do you think so