Answer:
Explanation:
1)
Given data:
Mass of lead = 25 g
Initial temperature = 40°C
Final temperature = 95°C
Cp = 0.0308 j/g.°C
Heat required = ?
Solution:
Specific heat capacity:
Cp
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 = Final temperature = initial temperature
ΔT = 95°C - 40°C
ΔT = 55°C
Q = 25 g × 0.0308 j/g.°C × 55°C
Q = 42.35 j
2)
Given data:
Mass = 3.1 g
Initial temperature = 20°C
Final temperature = 100°C
Cp = 0.385 j/g.°C
Heat required = ?
Solution:
Specific heat capacity:
Cp
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 = Final temperature = initial temperature
ΔT = 100°C - 20°C
ΔT = 80°C
Q = 3.1 g × 0.385 j/g.°C × 80°C
Q = 95.48 j
3)
Given data:
Mass of Al = ?
Initial temperature = 60°C
Final temperature = 30°C
Cp = 0.897 j/g.°C
Heat released = 120 j
Solution:
Specific heat capacity:
Cp
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 = Final temperature = initial temperature
ΔT = 30°C - 60°C
ΔT = -30°C
120 j = m × 0.897 j/g.°C × -30°C
120 j = m × -26.91 j/g
m = 120 j / -26.91 j/g
m = 4.46 g
negative sign show heat is released.
4)
Given data:
Mass of ice = 1.5 g
Change in temperature = ?
Cp = 0.502 j/g.°C
Heat added= 30.0 j
Solution:
Specific heat capacity:
Cp
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 = Final temperature = initial temperature
30.0 j = 1.5 g × 0.502 j/g.°C × ΔT
30.0 j = 0.753 j/°C × ΔT
30.0 j /0.753 j/°C = ΔT
39.84 °C = ΔT
