Question

If a 10.0 g piece of metal required 100 cal to raise its temperature
by 20°C, what would you report as its specific heat?
a. 10 cal/g·°C
b. 200 cal/g·°C
c. 0.5 cal/g·°C
d. 2 cal/g·°C

Answer 1

Answer:

C. 0.5 cal/g°C

Explanation:

Using the following equation:

Q = m × c × ∆T

Where;

Q = amount of heat absorbed/released (calories)

m = mass of substance (g)

c = specific heat capacity (calg/°C)

∆T = change of temperature (°C)

According to the information provided in this question;

m = 10g

c = ?

∆T = 20°C

Q = 100cal

Using Q = m × c × ∆T

c = Q ÷ (m × ∆T)

c = 100 ÷ (10 × 20)

c = 100 ÷ 200

c = 0.5 cal/g°C