(30)(Cm)Δ+(335.5)(4.18)Δ(21)=0
I need to find the specific heat of the masses, but i don't know how to, i'm currently doing a lab and these are the mass and temp. numbers that i plugged in...
To find the specific heat of the masses in the equation you provided, you can use the formula:
q = mcΔT
Where:
q is the amount of heat absorbed or released
m is the mass
c is the specific heat
ΔT is the change in temperature
In your equation, you have two terms involving different masses and temperature changes. Let's break it down:
(30)(Cm)Δ + (335.5)(4.18)Δ(21) = 0
The first term is (30)(Cm)Δ, where 30 is the mass, Δ represents the change in temperature, and Cm is the specific heat of the first mass.
The second term is (335.5)(4.18)Δ(21), where 335.5 is the mass, 4.18 is the specific heat of the second mass, and Δ(21) represents the change in temperature.
Since the equation is equal to zero, it implies that the total heat absorbed or released is zero. Therefore, you can equate the first and second term:
(30)(Cm)Δ = - (335.5)(4.18)Δ(21)
Now, you can solve for the specific heat (Cm):
Cm = - (335.5)(4.18)Δ(21) / (30)Δ
Simplifying this expression will give you the specific heat (Cm) of the masses involved in your lab experiment.