Question

What is the specific heat of a substance if a mass of 10.0 kg increases in temperature from 10.0°C to 70.0°C when 2,520 J of heat is applied?
Use q equals m C subscript p Delta T..

Answer 1

Plug in what you know
-Heat (q) is 2520
-Delta T (change in temperature) is 70-10 which is 60
-Mass (M) is 10kg
-specific heat (c) is unknown.

2520 = (10)C(60)
Use algebra to solve for C.
Multiply C by 60 and 10.
2520 = 600C
Divide both sides by 600 to get C by itself.
C = 4.2 J/g

Answer 2

Answer:

a

Explanation: