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ddd [48]
3 years ago
7

How much heat energy is required to raise the temperature of 0.358 kg of copper from 23.0 ∘C to 60.0 ∘C? The specific heat of co

pper is 0.0920 cal/(g⋅∘C) .
Chemistry
2 answers:
NNADVOKAT [17]3 years ago
7 0
In order to calculate how much heat is needed to raise the temperature you need to use the formula q =mass x specific heat x (final temperature- initial temperature) where q represents heat being absorbed or released. Before you begin you would convert kg to g because the specific heat is measure in g. So you would set up the equation as q = 358 g x .092 x (60-23 degrees Celsius) which would give you 1218.6
allsm [11]3 years ago
7 0

Answer:

1.22 × 10³ cal

Explanation:

Given data

  • Mass of Cu (m): 0.358 kg = 358 g
  • Initial temperature: 23.0°C
  • Final temperature: 60.0°C
  • Change in the temperature (ΔT): 60.0°C - 23.0°C = 37.0°C
  • Specific heat of Cu (c): 0.0920 cal/g.°C

We can calculate the heat (Q) required to raise the temperature of 0.358 kg of copper from 23.0°C to 60.0°C using the following expression.

Q = c × m × ΔT

Q = (0.0920 cal/g.°C) × 358 g × 37.0°C

Q = 1.22 × 10³ cal

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Two elements are studied. One has atomic number X and one has atomic number X+2. It is known that element X is a halogen. To wha
vladimir2022 [97]

Answer:

Group 1 (or IA)

Explanation:

If element X is a halogen, then it belongs to the group 17 (or VIIA, under a different notation).

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If an atom has three protons three neutrons and two electrons what is the electrical charge of the atom
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Answer:

Bueno,

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g A microwave oven heats by radiating food with microwave radiation, which is absorbed by the food and converted to heat. If the
Sliva [168]

Answer:

The total photons required = 5.19 × 10²⁸ photons

Explanation:

Given that:

the radiation wavelength λ= 12.5 cm = 0.125 m

Volume of the container = 0.250 L = 250 mL

The density of water = 1 g/mL

Density = mass /volume

Mass =  Volume ×  Density

Thus; the mass of the water =  250 mL ×  1 g/mL

the mass of the water = 250 g

the specific heat of water s = 4.18 J/g° C

the initial temperature T_1 = 20.0° C

the final temperature T_2 = 99° C

Change in temperature \Delta T = (99-20)° C = 79 ° C

The heat q absorbed during the process = ms \Delta T

The heat q absorbed during the process = 250 g × 4.18 J/g° C × 79° C

The heat q absorbed during the process = 82555 J

The energy of a photon can be represented by the equation :

= hc/λ

where;

h = planck's constant = 6.626 \times 10^{-34} \ J.s

c = velocity of light = 3.0 \times 10^8 \ m/s

=  \dfrac{6.626 \times 10^{-34} \times 3.0 \times 10^8}{0.125}

= 1.59024 \times 10^{-24} J

The total photons required = Total heat energy/ Energy of a photon

The total photons required = \dfrac{82555 J}{1.59024 \times 10^{-24}J}

The total photons required = 5.19 × 10²⁸ photons

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2 years ago
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