Question

How much energy is required to raise the temperature of a

300.0 gram block of lead from 22.3°C to 59.9°C? The specific
heat of lead is 0.129 J/gºC.
Round your answer to the nearest whole number.

Answer 1

Q = 1455 J

Step-by-step explanation:

Given data:

Mass of block = 300 g

Initial temperature = 22.3 °C

Final temperature = 59.9°C

Specific heat of lead = 0.129 j/g.°C

Energy required = ?

Solution:

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 = T2 - T1 = 59.9°C - 22.3 °C

ΔT = 37.6 °C

Q = 300 g × 0.129 j/g.°C × 37.6 °C

Q = 1455 J