if a 3.1g ring is heated using 10.0 calories, its temperatures rises 17.9C calculate the specific heat capacity of the ring
1 answer:
So what you do is you use the formula shown below:
specific heat capacity = energy required / (mass * change in temperature)
here ,
energy required = 10.0°C Note that cal. is short form for °C
mass (m) = 3.1g
change in temperature (ΔT) = 17.9°C Note that "ΔT" means change in temperature
So, plugging the values into the formula, we get,
Specific heat capacity=
=
= 0.1802 cal./g°C <span>i rounded the answer to the fourth decimal point
</span>
specific heat capacity = energy required / (mass * change in temperature)
here ,
energy required = 10.0°C Note that cal. is short form for °C
mass (m) = 3.1g
change in temperature (ΔT) = 17.9°C Note that "ΔT" means change in temperature
So, plugging the values into the formula, we get,
Specific heat capacity=
=
= 0.1802 cal./g°C <span>i rounded the answer to the fourth decimal point
</span>
You might be interested in
Answer:
4.9 m/s²
Explanation:
Draw a free body diagram. There are two forces on the object:
Weight force mg pulling straight down,
and normal force N pushing perpendicular to the plane.
Sum the forces in the parallel direction.
∑F = ma
mg sin θ = ma
a = g sin θ
a = (9.8 m/s²) (sin 30°)
a = 4.9 m/s²
Answer:
The speed of the electron is 1.371 x 10⁶ m/s.
Explanation:
Given;
wavelength of the ultraviolet light beam, λ = 130 nm = 130 x 10⁻⁹ m
the work function of the molybdenum surface, W₀ = 4.2 eV = 6.728 x 10⁻¹⁹ J
The energy of the incident light is given by;
E = hf
where;
h is Planck's constant = 6.626 x 10⁻³⁴ J/s
f = c / λ
Photo electric effect equation is given by;
E = W₀ + K.E
Where;
K.E is the kinetic energy of the emitted electron
K.E = E - W₀
K.E = 15.291 x 10⁻¹⁹ J - 6.728 x 10⁻¹⁹ J
K.E = 8.563 x 10⁻¹⁹ J
Kinetic energy of the emitted electron is given by;
K.E = ¹/₂mv²
where;
m is mass of the electron = 9.11 x 10⁻³¹ kg
v is the speed of the electron
Therefore, the speed of the electron is 1.371 x 10⁶ m/s.