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salantis [7]
3 years ago
8

While camping up at Haloooo Mountain (5,500 feet above sea level), it generally takes Sandy about the same amount of time to boi

l water for her soup as it takes to make a sandwich. This week, Sandy is camping at a trail that is only about 200 feet above sea level. Sandy is very hungry and wants something to eat immediately. Should she make the soup or sandwich? Which would be ready more quickly and why?
Chemistry
1 answer:
oksian1 [2.3K]3 years ago
6 0
She should make soup. When you are way above sea level, the atmospheric pressure decreases. Boiling starts when the vapor pressure of the liquid is the same with the atmospheric pressure. This would mean that water boils at a much lower temperature than the usual 100deg C. So it would take longer for the components in the soup to cook compared to when it is cooked on the foot of the mountain. 
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Explanation:

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5] Graphite is used as a good conductor of electricity.

6] Pottassium is used in fertilizers.

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3 0
3 years ago
A student constructs a coffee cup calorimeter and places 50.0 mL of water into it. After a brief period of stabilization, the te
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Answer:

The calorimeter constant is  = 447 J/°C

Explanation:

The heat absorbed or released (Q) by water can be calculated with the following expression:

Q = c × m × ΔT

where,

c is the specific heat

m is the mass

ΔT is the change in temperature

The water that is initially in the calorimeter (w₁) absorbs heat while the water that is added (w₂) later releases heat. The calorimeter also absorbs heat.

The heat absorbed by the calorimeter (Q) can be calculated with the following expression:

Q = C × ΔT

where,

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The density of water is 1.00 g/mL so 50.0 mL = 50.0 g. The sum of the heat absorbed and the heat released is equal to zero (conservation of energy).

Qabs + Qrel = 0

Qabs = - Qrel

Qcal + Qw₁ = - Qw₂

Qcal = - (Qw₂ + Qw₁)

Ccal . ΔTcal = - (cw . mw₁ . ΔTw₁ + cw . mw₂ . ΔTw₂)

Ccal . (30.31°C - 22.6°C) = - [(4.184 J/g.°C) × 50.0 g × (30.31°C - 22.6°C) +  (4.184 J/g.°C) × 50.0 g × (30.31°C - 54.5°C)]

Ccal  = 447 J/°C

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3 years ago
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