Question

When 386 J of heat is added to 3.2 g of a substance at 23 degrees Celsius, the temperature increases to 79 degrees Celsius. What is the specific heat of the substance with the correct units of measure?

Answer 1

The specific heat of the substance is 2.15 J/g⁰C

Step-by-step explanation

Given:

Quantity of heat, Q = 386 J

Mass of the substance, m = 3.2 g

Initial temperature, T₁ = 23 ⁰C

Final temperature, T₂ = 79 ⁰C

What to find:

The specific heat of the substance.

Step-by-step solution:

The specific heat of the substance, c can be calculated using the formula below.

`Q=mc\Delta T`

ΔT = T₂ - T₁ = 79 ⁰C - 23 ⁰C = 56 ⁰C

So plugging Q = 386 J, m = 3.2 g and ΔT = 56 ⁰C into the formula, we have

`\begin{gathered} 386J=3.2g* c*56^0C \\ \\ c=(386J)/(3.2g*56^0C)=2.15\text{ }J\text{/}g^0C \end{gathered}`

The specific heat of the substance is 2.15 J/g⁰C