A 6.0g pellet of lead at 32 degrees c gains 36.8 J of heat. What will be its final temperature
The temperature RISE will be Q/(mC)
where m is the mass and C is the specific heat of lead. You will need to look that up.
To find the final temperature of the lead pellet, you can use the equation for heat transfer:
q = mcΔT
Where:
q is the heat transferred
m is the mass of the object (in this case, the lead pellet)
c is the specific heat capacity of the material (in this case, lead)
ΔT is the change in temperature
In the given problem, we are given:
q = 36.8 J (heat transferred)
m = 6.0 g (mass of lead pellet)
To find the specific heat capacity of lead, you can refer to a reliable source or use an average value. The specific heat capacity of lead is approximately 0.13 J/g°C.
Plugging in the values into the equation, we get:
36.8 J = (6.0 g) * (0.13 J/g°C) * ΔT
Now, we need to solve for ΔT (change in temperature). To do this, rearrange the equation:
ΔT = 36.8 J / [(6.0 g) * (0.13 J/g°C)]
Calculate the value of ΔT using the provided numbers:
ΔT = 36.8 J / [(6.0 g) * (0.13 J/g°C)]
ΔT = 36.8 J / 0.78 J/°C
ΔT ≈ 47.2°C
Finally, to find the final temperature, add the change in temperature to the initial temperature:
Final temperature = Initial temperature + ΔT
Final temperature = 32°C + 47.2°C
Final temperature ≈ 79.2°C
Therefore, the final temperature of the lead pellet will be approximately 79.2°C.