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
