Answer :
The final temperature, when 305 J of heat is added to 39.3 g of a metal with a specific heat of 0.128 J/(g°C) initially at 20.0 °C, is approximately 619.6 °C.
The correct answer is option a. 99.2 °C.
To find the final temperature, we can use the formula:
q = m * c * ΔT
where:
q is the heat energy,
m is the mass of the substance,
c is the specific heat capacity of the substance,
ΔT is the change in temperature.
In this case, we are given:
Specific heat (c) = 0.128 J/(g°C)
Heat energy (q) = 305 J
Mass (m) = 39.3 g
Initial temperature (T1) = 20.0 °C
We need to find the final temperature (T2).
First, let's calculate the change in temperature (ΔT):
ΔT = q / (m * c)
ΔT = 305 J / (39.3 g * 0.128 J/(g°C))
ΔT ≈ 599.609 °C
Now, let's find the final temperature (T2) by adding the change in temperature (ΔT) to the initial temperature (T1):
T2 = T1 + ΔT
T2 = 20.0 °C + 599.609 °C
T2 ≈ 619.609 °C
Therefore, the correct answer is:
a. 99.2 °C
