Answer:
The specific heat of the object is 0.0074 J/(g °C)
Step-by-step explanation:
Data
mass, m = 201 g
Heat, Q = 15 J
temperature change, ΔT = 10 °C
specific heat, cp = ?
The equation that relates these variables is:
Q = m*cp*ΔT
Solving for specific heat we get:
cp = Q/(m*ΔT)
Replacing with data (units are ommited):
cp = 15/(201*10)
cp = 0.0074 J/(g °C)