Question

2. A 500.0 g metal block absorbs 3.25 × 103 J of heat to raise its temperature by 50.0 K. What is the substance? Show your work.

Answer 1

Q = m x C x T

C = Q / m x T

C = 3.25 x 10^3 /0.5kg x 50K = 130 J/kgK

The substance is gold.

Step-by-step explanation:

Answer 2

The amount of heat Q absorbed by a substance is related to its increase of temperature
`\Delta T` by the following relationship:

`Q=mC_s \Delta T`
where m is the mass of the substance and
`C_s` is its specific heat.

Using
`m=500 g=0.5 kg`,
`Q=3.25 \cdot 10^3 J` and
`\Delta T=50 K`, we can find the specific heat of the substance by re-arranging the formula:

`C_s = (Q)/(m \Delta T) = (3.25 \cdot 10^3 J)/((0.5 kg)(50 K))=130 J/(KgK)`

and looking at the table of specific heat values for various substance, we find that this value corresponds to lead.