Answer:
6.36 105 J
Explanation:
In calorimetry all the energy given to a system is converted to heat and the equation for heat is
Q = m ΔT
The temperature can be in degrees Celsius or Kelvin since the interval between them is the same, substitute and calculate
Q = 1.9 4186 (373-293)
Q = 6.36 105 J
This heat equals the energy supplied
Answer:
338 K
Explanation:
= 2.5 kg, c = 4189 J/(kg K), = 13.5 , = 22.5
= 0.5 kg, = 20
Heat loss by hotter water = heat gained by cooler water
cΔT = cΔT
c( - ) = c( - )
2.5 x 4189 x (22.5 - 13.5) = 0.5 x 4189 x ( - 20)
2.5 x 4189 x 9 = 2094.5 ( - 20)
94252.5 = 2094.5 - 41890
94252.5 + 41890 = 2094.5
136142.5 = 2094.5
=
= 65
= 65
But,
θ K = 273 + θ
= 273 + 65
= 338
The final temperature of the water is 338 K.
Answer:
x_total = 0.17m
Explanation:
We can treat this exercise with the kinematics equations, where in the first part it is accelerated and in the second it is a uniform movement.
Let's analyze accelerated motion
The time that lasts is t = 20 10⁻³ s, the initial speed is zero (v₀ = 0), let's find the length that advances
x₁ = v₀ t + ½ a t²
x₁ = ½ a t²
x₁ = ½ 210 (20 10⁻³)²
x₁ = 4.2 10⁻² m
let's find the speed for the end of this movement
v = v₀ + a t
v = 0 + 210 20 10⁻³
v = 4.2 m / s
with this speed we can find the distance that the uniform movement
x₂ = v t2
x₂ = 4.2 30 10⁻³
x₂ = 1.26 10⁻¹ m
x₂ = 0.126m
the total distance traveled is
x_total = x₁ + x₂
x_total = 0.0420 +0.126
x_total = 0.168m
Let's reduce the significant figures to two
x_total = 0.17m
Answer:
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Explanation:
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