The new temperature of the piece of lead will be 37.5°C
SPECIFIC HEAT CAPACITY:
Q = m × c × ∆T
Where;
Q = quantity of heat absorbed or released (J)
m = mass of substance (g)
c = specific heat capacity
∆T = change in temperature (°C)
According to this question,
Q = 1.90J
m = 1.29g
T1 = 26°C
T2 = ?
c = 0.128 J/g°C
1.90 = 1.29 × 0.128 × (T2- 26°C)
1.90 = 0.165 (T2 - 26°C)
1.90 = 0.165T2 - 4.29
1.90 + 4.29 = 0.165T2
6.19 = 0.165T2
T2 = 37.5°C
Therefore, the new temperature of the piece of lead will be 37.5°C.
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