Question

How many joules (J) are needed to increase the temperature of 10.0 g of lead (Pb) from 30 ˚C to 50 ˚C? (Cp for Pb = 0.128 J/g˚C)

Answer 1

Answer:

Q = 25.6 j

Explanation:

Given data:

Energy needed= ?

Mass of lead = 10.0 g

Initial temperature = 30 °C

Final temperature = 50°C

Cp = 0.128 j/g.°C

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 50°C - 30°C

ΔT = 20°C

Now we will put the values in formula.

Q = 10 g × 0.128 j/g.°C × 20°C

Q = 25.6 j