When we cook a marshmallow on a metal poker tool over an open flame, there are three ways in which heat energy is transferred: Conduction, convection, and radiation.
<h3>Heat energy transfer</h3>
Heat transfer is the natural transfer of heat from an object with a higher temperature to an object with a lower temperature. Heat transfer can occur in three ways, namely conduction, convection, and radiation.
- Conduction occurs when heat flows from a place with a high temperature to a place with a lower temperature using a fixed heat-conducting medium. Heat transfer from the open flame to the marshmallows via direct fire contact with the marshmallows is an example of conduction.
- Convection is the transfer of heat by means of a stream in which the intermediate substance also moves. If the particles move and cause heat to propagate, convection will occur. The hot air rising from the flames burning the marshmallows is an example of convection.
- Radiation is heat transfer without a medium. Radiation can also usually be accompanied by light. The direct transfer of heat from the flame to the marshmallow in the form of waves is an example of radiation.
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The change in the player's internal energy is -491.6 kJ. The number of nutritional calories is -117.44 kCal
For this process to take place, some of the basketball player's perspiration must escape from the skin. This is because sweating relies on a physical phenomenon known as the heat of vaporization.
The heat of vaporization refers to the amount of heat required to convert 1g of a liquid into a vapor without causing the liquid's temperature to increase.
From the given information,
- the work done on the basketball is dW = 2.43 × 10⁵ J
The amount of heat loss is represented by dQ.
where;
∴
Using the first law of thermodynamics:b
dU = dQ - dW
dU = -mL - dW
dU = -(0.110 kg × 2.26 × 10⁶ J/kg - 2.43 × 10⁵ J)
dU = -491.6 × 10³ J
dU = -491.6 kJ
The number of nutritional calories the player has converted to work and heat can be determined by using the relation:
dU = -117.44 kcal
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Answer:
The final charges of each sphere are: q_A = 3/8 Q
, q_B = 3/8 Q
, q_C = 3/4 Q
Explanation:
This problem asks for the final charge of each sphere, for this we must use that the charge is distributed evenly over a metal surface.
Let's start Sphere A makes contact with sphere B, whereby each one ends with half of the initial charge, at this point
q_A = Q / 2
q_B = Q / 2
Now sphere A touches sphere C, ending with half the charge
q_A = ½ (Q / 2) = ¼ Q
q_B = ¼ Q
Now the sphere A that has Q / 4 of the initial charge is put in contact with the sphere B that has Q / 2 of the initial charge, the total charge is the sum of the charge
q = Q / 4 + Q / 2 = ¾ Q
This is the charge distributed between the two spheres, sphere A is 3/8 Q and sphere B is 3/8 Q
q_A = 3/8 Q
q_B = 3/8 Q
The final charges of each sphere are:
q_A = 3/8 Q
q_B = 3/8 Q
q_C = 3/4 Q
<span>Δ</span>E = q + w
q = heat (quantity of)
q and w can be positive or negative depending on if work/heat is being absorbed/done on the system or released/done by the system
