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Inessa [10]
3 years ago
10

Three people are gathered around a campfire. One has his hands cupped around a ceramic mug of hot chocolate to warm them. Anothe

r is toasting a marshmallow above the fire. The third is roasting a hot dog above the glowing coals. Identify the primary source of heat transfer each person is
enjoying
Chemistry
2 answers:
NeX [460]3 years ago
8 0

Answer:

Person one has conduction, person 2 has radiation, and person 3 has convention.

Explanation:

Person 1 is touching the mug to get warm, which is a transfer of heat.

Person 2 is exposed to the fire, which is radiation.

Person 3 is exposed to the warm air of convection.

luda_lava [24]3 years ago
6 0

Explanation:

A process that involves transfer of heat from a hot substance to a cold substance by coming in contact with each other is known as conduction.

For example, a person has his hands cupped around a ceramic mug of hot chocolate then heat is transferring from ceramic mug to the hands.

Whereas a process in which heat energy is transferred in a wave-like motion through the space is known as radiation.

For example, a person is toasting a marshmallow above the fire is getting heat energy in the form of radiation.

Also, third person who is roasting a hot dog above the glowing coals is getting heat energy in the form of radiation.

Hence, we can conclude that primary source of heat transfer for person 1 is conduction, and for both person 2 and 3 is radiation.

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anzhelika [568]

Answer:

\boxed{\text{25. 20 L; 26. 49 K}}

Explanation:

25. Boyle's Law

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Data:

\begin{array}{rcrrcl}p_{1}& =& \text{100 kPa}\qquad & V_{1} &= & \text{10.00 L} \\p_{2}& =& \text{50 kPa}\qquad & V_{2} &= & ?\\\end{array}

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\begin{array}{rcl}100 \times 10.00 & =& 50V_{2}\\1000 & = & 50V_{2}\\V_{2} & = &\textbf{20 L}\\\end{array}\\\text{The new volume will be } \boxed{\textbf{20 L}}

26. Ideal Gas Law

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V = 20 L

n = 5 mol

R = 8.314 kPa·L·K⁻¹mol⁻¹

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101.3 × 20 = 5 ×  8.314 × T

2026 = 41.57T

T = \dfrac{2026}{41.57} = \textbf{49 K}\\\\\text{The Kelvin temperature is }\boxed{\textbf{49 K}}

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