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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?

Respuesta :

meerkat18
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. 
Edufirst

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

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