A 2.00-kg metal object requires 1.00 × 10⁴ J of heat to raise its temperature from 20.0°C to 40.0°C. What is the specific heat capacity of the metal? A)50.0 J/(kg · C°) B)125 J/(kg · C°) C)250 J/(kg · C°) D)500 J/(kg · C°) E)1.00 × 10³ J/(kg · C°)

Question

Quizplus · Accepted Answer

We can use the formula for heat absorbed or released by an object: Q = mcΔT where Q is the heat absorbed or released, m is the mass of the object, c is the specific heat capacity of the material, and ΔT is the change in temperature. Plugging in the given values, we get: 1.00 × 10⁴ J = (2.00 kg)(c)(40.0°C - 20.0°C) Simplifying, we get: c = (1.00 × 10⁴ J)/(2.00 kg)(20.0°C) c ≈ 250 J/(kg · C°) Therefore, the specific heat capacity of the metal is 250 J/(kg · C°), which is answer choice C).

A 200-Kg Metal Object Requires 1