Heat required to melt 0.05 kg of aluminum is 28.7 kJ.
<h3>What is the energy required to melt 0.05 kg of aluminum?</h3>
The heat energy required to melt 0.05 kg of aluminum is obtained from the heat capacity of aluminum and the melting point of aluminum.
The formula to be used is given below:
- Heat required = mass * heat capacity * temperature change
Assuming the aluminum sheet was at room temperature initially.;
Room temperature = 25 °C
Melting point of aluminum = 660.3 °C
Temperature difference = (660.3 - 25) = 635.3 903
Heat capacity of aluminum = 903 J/kg/903
Heat required = 0.05 * 903 * 635.3
Heat required = 28.7 kJ
In conclusion, the heat required is obtained from the heat change aluminum and the mass of the aluminum melted.
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Answer:
that best describes the process is C
Explanation:
This problem is a calorimeter process where the heat given off by one body is equal to the heat absorbed by the other.
Heat absorbed by the smallest container
Q_c = m ce (
-T₀)
Heat released by the largest container is
Q_a = M ce (T_{i}-T_{f})
how
Q_c = Q_a
m (T_{f}-T₀) = M (T_{i} - T_{f})
Therefore, we see that the smaller container has less thermal energy and when placed in contact with the larger one, it absorbs part of the heat from it until the thermal energy of the two containers is the same.
Of the final statements, the one that best describes the process is C
since it talks about the thermal energy and the heat that is transferred in the process
Answer:
The displacement in t = 0,
y (0) = - 0.18 m
Explanation:
Given f = 40 Hz , A = 0.25m , μ = 0.02 kg / m, T = 20.48 N
v = √ T / μ
v = √20.48 N / 0.02 kg /m = 32 m/s
λ = v / f
λ = 32 m/s / 40 Hz = 0.8
K = 2 π / λ
K = 2π / 0.8 = 7.854
φ = X * 360 / λ
φ = 0.5 * 360 / 0.8 = 225 °
Using the model of y' displacement
y (t) = A* sin ( w * t - φ )
When t = 0
y (0) = 0.25 m *sin ( w*(0) - 225 )
y (0) = 0.25 * -0.707
y (0) = - 0.18 m
