A 1.0 kg object absorbs 1,303 J of heat energy and experiences a temperature increase of 5.2∘C. What is the object’s specific heat, in joules per gram-degree celsius? Report your answer with the correct number of significant figures.

c = 250.58 J/kg/

Explanation:

The specific heat of a substance is the required quantity of heat to increase or decrease the temperature of its unit mas by 1 kelvin.

Q = mcΔθ

where: Q is the quantity of heat absorbed or released, m is the mass of the substance, c is its specific heat and Δθ is the change in temperature of the substance.

Given that; m = 1.0 kg, Q = 1303 J and Δθ = 5.2 , then;

c = Q ÷ (mΔθ)

  = 1303 ÷ (1.0 × 5.2)

  = 1303 ÷ 5.2

  = 250.58 J/kg/

The specific heat of the object is 250.58 J/kg/.

have a great day because if you dont you and i will be sad