High School

A 2.00 kg metal object requires \(5.02 \times 10^3 \text{ J}\) of heat to raise its temperature from \(20.0^\circ\text{C}\) to \(40.0^\circ\text{C}\). What is the specific heat capacity of the metal?

A. 63.0 J/(kg·°C)
B. 126 J/(kg·°C)
C. 251 J/(kg·°C)
D. 502 J/(kg·°C)

Answer :

The specific heat capacity of the metal object is b. 126 J/(kg·°C).

To determine the specific heat capacity (c), we use the formula:

Q = mcΔT

where Q is the heat added (5020 J), m is the mass (2.00 kg), and ΔT is the change in temperature (40.0°C - 20.0°C = 20.0°C).

Rearranging the formula to solve for c:

c = Q / (mΔT)

Substitute the known values into the equation:

c = 5020 J / (2.00 kg * 20.0°C)

c = 5020 J / 40(kg·°C)

c = 125.5 J/(kg·°C)

c = 126 J/(kg·°C)

Therefore, the specific heat capacity of the metal is 126 J/(kg·°C).

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