Question

What is the heat capacity of a substance in which 1050 grams will rise in temperature from 30.4°C to 25.7°C when it absorbs 2880 joules of heat?

Answer 1

0.584 J/g°C

Step-by-step explanation:

Using the formula as follows;

Q = m × c × ∆T

Where;

Q = amount of heat (joules)

m = mass of substance (g)

c = specific heat capacity (J/g°C)

∆T = change in temperature (°C)

According to the information in this question;

mass of substance (m) = 1050g

Initial temperature = 25.7°C

Final temperature = 30.4°C

amount of heat (Q) = 2880J

change in temperature (∆T) = (30.4°C - 25.7°C) = 4.7°C

Using Q = m.c.∆T

c = Q ÷ m∆T

c = 2880 ÷ (1050 × 4.7)

c = 2880 ÷ 4935

c = 0.584 J/g°C

The specific capacity of the substance is 0.584 J/g°C.