A lead fishing weight with a mass of 57g absorbs 146 cal of heat. If its initial temperature is 47 degrees Celsius, what is its final temperature?
The heat and the corresponding difference in the temperature of the system can be related through the equation,
H = mcp(ΔT)
where H is heat, m is mass, cp is specific heat and equal to 1 cal/g°C for water, and ΔT is the temperature difference equal to T2 - T1. Substituting the known values,
146 cal = (57 g)(1 cal/ g°C)(T2 - 47°C)
The value of T2 from the equation is approximately 49.56°C.
Answer: 130°C
Explanation:
1) Data:
i) material: lead
ii) m = 57 g
iii) Q = 146 cal
iv) Ti = 47°C
v) Tf = ?
2) Formulas
Q = m C (Tf - Ti)
C is the specific heat of lead. You must find that value in tables using your textbook or researching in internet.
Such value is C = 0.031 cal / g°C
3) Solution
Q = m C (Tf - Ti) ⇒ Tf - Ti = Q / m C ⇒ Tf = Ti + Q /m C
⇒Tf = 47°C + 146 cal / (57g × 0.031 cal / g°C) = 47°C + 82.6°C = 129.6°C ≈ 130°C