A lead fishing weight with a mass of 57 g absorbs 146 cal of heat.If its initial temperature is 48^C, what is its final temperature?
To find the final temperature of the lead fishing weight, we can use the equation:
Q = mcΔT
Where:
Q = heat absorbed/released (in calories),
m = mass (in grams),
c = specific heat capacity (in cal/g°C),
ΔT = change in temperature (in °C).
Given:
Q = 146 cal
m = 57 g
Initial temperature = 48°C
First, let's determine the specific heat capacity of lead. The specific heat capacity of lead is approximately 0.03 cal/g°C.
Now let's rearrange the equation to solve for ΔT:
ΔT = Q / (mc)
Substituting the given values:
ΔT = 146 cal / (57 g * 0.03 cal/g°C)
ΔT = 146 cal / 1.71 cal/°C
ΔT ≈ 85.38 °C
To find the final temperature, we add the change in temperature to the initial temperature:
Final temperature = Initial temperature + ΔT
Final temperature = 48°C + 85.38 °C
Final temperature ≈ 133.38 °C
Therefore, the final temperature of the lead fishing weight is approximately 133.38°C.
To find the final temperature of the lead fishing weight, we can use the equation:
Q = mcΔT
where:
Q is the heat absorbed (in calories),
m is the mass (in grams),
c is the specific heat capacity of the substance (in cal/g°C),
ΔT is the change in temperature (in °C).
Given:
Q = 146 cal
m = 57 g
c = specific heat capacity of lead (0.03 cal/g°C)
Initial temperature, Ti = 48°C
Final temperature, Tf = ?
We can rearrange the equation to solve for ΔT:
Q = mcΔT -> ΔT = Q / (mc)
Substituting the given values:
ΔT = 146 cal / (57 g × 0.03 cal/g°C)
Now, we can calculate ΔT:
ΔT = 146 / (1.71)
≈ 85.38 °C
Finally, we can find the final temperature, Tf, by adding ΔT to the initial temperature, Ti:
Tf = Ti + ΔT
= 48 + 85.38
≈ 133.38 °C
Therefore, the final temperature of the lead fishing weight is approximately 133.38 °C.
57
Q=mc∆t
Q = heat energy
m = mass
c = specific heat (of lead) [you need to look this number up]
∆t= change in temperature (= final - initial temp)