Question

What is the specific heat of a 3.78 kg object that absorbs 678 J as the temperature increases by 4.25 K?

Answer 1

Below

Step-by-step explanation:

Specific heat has units of J / (kg C)

( a degree of C is the same as one K)

Sub in the values

678 J / (3.78 kg * 4.25 C) = spec heat = 42.2 J/(kg-K)