Answer:
Explanation:
1)
Given data:
Heat required = ?
Mass of sample = 225 g
Initial temperature = 25°C
Final temperature = 60°C
Specific heat capacity of lead = 0.128 j/g°C
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 = 60°C - 25°C
ΔT = 35°C
Q = 225 g × 0.128 j/g°C × 35°C
Q = 1008 j
2)
Given data:
Heat capacity for 1°C = 222 j
Mass of sample = 246 g
Specific heat capacity of sample = ?
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
222 j = 246 g × c × 1°C
c = 222 j / 246 g.°C
c = 0.902 j/g.°C
The given metal is aluminium.
