Question

A 300 g block of a substance requires 2340 J of heat to raise its temperature from 25°C to 45°C.

What is the substance?

Use the equation q = mCΔT and then use the table to identify which substance the block is.

Copper

Gold

Ice

Aluminum

Answer 1

Mass of the block = 300g

Heat required = 2340 J

Change in temp. = (45 -25) = 20

According to the formula : q = m C (t2 - t1)

2340 = 300/1000 C 20

C = 2340/20 * 1000/300

C = 2340/6 = 390

Answer 2

The substance in the block is copper.

To identify the substance, we can use the following steps:

1. Calculate the specific heat capacity of the substance using the equation:`

`C = q / mΔT`

where:

* C is the specific heat capacity
`(J/g°C)`

* q is the amount of heat absorbed (J)

* m is the mass of the substance (g)

* ΔT is the change in temperature (°C)

Substituting the known values, we get:

C = 2340 J / 300 g * (45°C - 25°C)

= 0.39
`J/g°C`

2. Look up the specific heat capacity of different substances in the table provided.

3. The substance with the specific heat capacity closest to 0.39
`J/g°C`is copper.

Therefore, the substance in the block is copper.

Answer: Copper