Question

A block of metal weighing 238 g is cooled from 128.7°C to 30.2°C with the release of 4850 joules of heat. Calculate the specific heat of the metal.

a) 0.25 J/g°C
b) 0.50 J/g°C
c) 0.75 J/g°C
d) 1.00 J/g°C

Answer 1

To find the specific heat of the metal, the formula q = mcΔT is used where q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. Plugging in the values and calculating, the answer closest to the calculated specific heat is Option a) 0.25 J/g°C.

Step-by-step explanation:

To calculate the specific heat of the metal, we can use the formula:

q = mcΔT where:

• q is the amount of heat in joules.
• m is the mass of the substance in grams.
• c is the specific heat capacity in J/g°C.
• ΔT is the change in temperature in °C.

In this case, the metal releases 4850 joules of heat (q = -4850 J), has a mass of 238 g (m), and experiences a temperature change (ΔT) from 128.7°C to 30.2°C, which is a decrease of 98.5°C.

Using the formula:
-4850 J = (238 g)(c)(98.5°C)

Solving for c, the specific heat capacity, we get:

c = -4850 J / (238 g • 98.5°C)

c = -4850 J / (23443 g°C)

c = 0.207 J/g°C

The closest answer from the options is Option a) 0.25 J/g°C.