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Juli2301 [7.4K]
3 years ago
12

How much heat must be removed from 456 g of water at 25. 0 c to hcange it into ice at 10 c

Chemistry
1 answer:
ValentinkaMS [17]3 years ago
6 0

The heat required to convert water is given as the product of the mas and the latent heat. 28.34 kJ heat must be removed from the water to change it into ice.

<h3>What is heat energy?</h3>

Heat energy is the product of mass, specific heat capacity, and temperature change. It is given as,

\rm Q = mc \Delta T

Given,

Mass of water = 456 gm

Specific heat capacity = 4.186 J / g K

Temperatutre change = 14.85 K

Substituting values above:

\begin{aligned} \rm Q &= 456 \times 4.186 \times 14.85\\\\&= 28345.91\;\rm J\\\\&= 28.34\;\rm kJ\end{aligned}

Therefore, 28.34 kJ of heat energy should be removed from the water.

Learn more about the heat here:

brainly.com/question/14052023

#SPJ4

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How many moles of carbon are 8.7x1023 atoms of carbon?
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Butoxors [25]

Answer:

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Explanation:

Mainly because a solution is a mixture of chemically bonded elements.

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Mercury melts at 234.3 k. Convert this this melting point to degrees celcius
DerKrebs [107]

Answer:

2069.85°C

Explanation:

0 K − 273.15 = -273.1 °C

Then,

234 K − 273.15 = 2069.85 °C

Best regards

3 0
4 years ago
Read 2 more answers
determine the final temperature of a gold nugget parentheses Mass equals 376 G parentheses that starts at 398k and loses 4.85 KJ
maw [93]

Answer:

297.23 K

Explanation:

Given that:

the mass of the gold nugget (m) = 376 g

the initial temperature T_1 = 398 K

amount of heat lost Q = -4.85 kJ = -4.85 × 10³

specific heat capacity (c) = 0.128 J/g° C

Using the formula for calculating the Heat energy

Q = mc\Delta T \\  \\  Q = m\times c\times (T_2-T_1)

-4.85 \times 10^3 = 376 \times  0.128 \times (T_2 - 398) \\ \\ -4.85\times 10^3 =  48.128(T_2 - 398) \\ \\ \dfrac{-4.85\times 10^3}{48.128}= (T_2 - 398) \\ \\-100.77 = T_2 -398\\\\T_2 = -398 +100.77\\ \\ \mathbf{ T_2 = 297.23K}

7 0
3 years ago
What is the frequency and energy per quantum (in Joules) of :
viktelen [127]

(a) f = 5.00 × 10²⁰ Hz, E = 3.32 × 10⁻¹³ J;

(b) f = 1.20 × 10¹⁰ Hz, E = 7.96 × 10⁻²⁴J.

<h3>Explanation</h3>

What's the similarity between a gamma ray and a microwave?

Both gamma rays and microwave rays are electromagnetic radiations. Both travel at the speed of light at 3.00 \times 10^{8}\;\text{m}\cdot\text{s}^{-1} in vacuum.

f = \dfrac{c}{\lambda}

where

  • f is the frequency of the electromagnetic radiation,
  • c is the speed of light, and
  • \lambda is the wavelength of the radiation.

(a)

Convert all units to standard ones.

\lambda = 0.600\;\text{pm} = 0.600 \times 10^{-12} \;\text{m}.

The unit of f shall also be standard.

f = \dfrac{c}{\lambda} = \dfrac{3.00\times 10^{8}\;\text{m}\cdot\text{s}^{-1}}{0.600\times 10^{12}\;\text{m}} = 5.00 \times 10^{20}\;\text{s}^{-1}= 5.00\times 10^{20}\;\text{Hz}.

For each particle,

E = h\cdot f,

where

  • E is the energy of the particle,
  • h is the planck's constant where h = 6.63\times 10^{-34}\;\text{J}\cdot\text{s}^{-1}, and
  • f is the frequency of the particle.

E = h \cdot f = 6.63\times10^{-34}\;\text{J}\cdot\text{s}\times 5.00\times 10^{20}\;\text{s}^{-1} = 3.32\times10^{-13}\;\text{J}.

(b)

Try the steps in (a) for this beam of microwave with

  • \lambda = 2.50 \;\text{cm} = 2.50\times 10^{-2}\;\text{m}.

Expect the following results:

  • f = 1.20\times 10^{10}\;\text{Hz}, and
  • E = 7.96\times 10^{-24}\;\text{J}.
4 0
3 years ago
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