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3 years ago

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

1 answer:

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

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Mercury melts at 234.3 k. Convert this this melting point to degrees celcius

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Answer:

2069.85°C

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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

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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

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

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

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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.