a 32g sample of copper is at 29C If 1502j of energy are added to the copper, what is the final temperature? Assume the specific heat of copper is 387J/KG C
Q=cm(t₂-t₁)
Q= 1502 J,
m=0.032kg,
t₁=29℃
c=387 J/kg•℃
Solve for t₂
To find the final temperature of the copper, we can use the formula for heat:
Q = mcΔT
Where:
Q = heat energy (in joules)
m = mass of the substance (in kilograms)
c = specific heat capacity (in joules per kilogram per Celsius)
ΔT = change in temperature (in Celsius)
In this case, we are given:
Q = 1502 J
m = 32 g = 0.032 kg (since specific heat is given in joules per kilogram per Celsius)
c = 387 J/kg°C
The initial temperature, Ti = 29°C
Let's calculate the change in temperature (ΔT) using the formula:
ΔT = Q / (mc)
Substituting the given values:
ΔT = 1502 J / (0.032 kg * 387 J/kg°C)
ΔT ≈ 12.30°C
To find the final temperature, we add the change in temperature to the initial temperature:
Tf = Ti + ΔT
Tf = 29°C + 12.30°C
Tf ≈ 41.30°C
Therefore, the final temperature of the copper sample is approximately 41.30°C.