The mass of an object is 201 g The object requires 150J to raise its temperature by 100°C. What is the specific heat of the object? 7.46 J/kg-K 0.00130 J/kg - K 3,020,000 J/kg . K 1500 J/kg . K​

Question

The mass of an object is 201 g The object requires 150J to raise its temperature by 100°C. What is the specific heat of the object?

7.46 J/kg-K

0.00130 J/kg - K

3,020,000 J/kg . K

1500 J/kg . K​

Answer 1

Q = mc_p(T₂-T₁) where Q = heat, m = mass, c_p = heat capacity (specific heat)

(T₂-T₁) = temperature rise

Units are Joules = Kg(J/(Kg-C))(C)

In this case 150 = (.201)(c_p)(100) so, c_p = 150/(.201*100) = 7.46 J/(Kg-C)