All categories

3 years ago

How much heat do you need to raise the temperature of 100g of aluminum from 30 C to 150 C

Chemistry

2 answers:

marysya [2.9K]3 years ago

5 0

Answer:

The Answer is actually 10.8 kJ

Explanation:

It's not -10.8 KJ y'all, do NOT get confused. I got it wrong when I wrote -10.8 KJ. I took the test and it's 10.8 KJ without the negative.

SVEN [57.7K]3 years ago

4 0

Answer:

Q = 10.8 KJ

Explanation:

Given data:

Mass of Al= 100g

Initial temperature = 30°C

Final temperature = 150°C

Heat required = ?

Solution:

Specific heat of Al = 0.90 j/g.°C

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 150°C - 30°C

ΔT = 120°C

Q = 100g×0.90 J/g.°C× 120°C

Q = 10800 J (10800j×1KJ/1000 j)

Q = 10.8 KJ

You might be interested in

Which group of elements readily loses two electrons to form a compound?

nexus9112 [7]

The alkaline earth metals (the second group) because their ion charge is +2

8 0

4 years ago

at a particular temperature, asample of pure water has a Kw of 1.7x10-12. what is the hydroxide concentration of this sample

Lena [83]

Answer:

Hydroxide concentration of the sample is 1.3x10⁻⁶M

Explanation:

The equilibrium constant of water, Kw, is:

H₂O(l) ⇄ H⁺(aq) + OH⁻(aq)

Kw is defined as:

Kw = 1.7x10⁻¹² = [H⁺] [OH⁻]

As the sample is of pure water, both H⁺ and OH⁻ ions have the same concentration because come from the same equilibrium, that is:

[H⁺] = [OH⁻]

We can write the Kw expression:

1.7x10⁻¹² = [OH⁻] [OH⁻]

1.7x10⁻¹² = [OH⁻]²

1.3x10⁻⁶M = [OH⁻]

<h3>Hydroxide concentration of the sample is 1.3x10⁻⁶M</h3>

6 0

3 years ago

Which change is exothermic?

Harlamova29_29 [7]

It’s the last one icebergs shrink as global temp rises

5 0

4 years ago

Read 2 more answers

How many liters of CH4(g), measured at 22.6 ∘C and 768 mmHg , must be burned to provide the heat needed to vaporize 3.82 L of wa

Dovator [93]

For H2O(l) at 100 ∘C, d=0.958gcm−3 and ΔvapH=40.7kJmol−1.

7 0

3 years ago

A sample of brass is put into a calorimeter (see sketch at right) that contains of water. The brass sample starts off at and the

Ierofanga [76]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The specific heat is $c_b = 0.402 J / g \cdot ^oC$

Explanation:

From the question we are told that

The mass of the sample is $m = 54.4 \ g$

The mass of the water is $m_w = 150.0 \ g$

The initial temperature of the sample is $T_i = 95.1 ^oC$

The initial temperature of the water is $T_{w_i} = 15^oC$

The final temperature of the water is $T = 17.6 ^oC$

Note the final temperature of water is equal to the final temperature of brass sample

The pressure is $P =1 \ atm$

Generally for according to the law of energy conservation

The heat lost by sample = The heat gain by water

The heat lost by brass sample is mathematically evaluated as

$H_L = m * c_b * [T_i - T]$

Where $c_b$ is the specific neat of the brass sample

The heat gained by water is mathematically evaluated as

$H_g = m_w *c_w * [T_w - T ]$

where $c_w$ is the specific heat of water which has a constant value of

$c_w = 4.186 joule/gram$

$H_L = H_g \ \equiv m* c_b * [T_i -T] = m_w * c_w * [T - T_w]$

substituting values

$52.4 * c_b * [95.1 - 17.6] = 150 * 4.186 * [ 17.6 - 15.0]$

$c_b = 0.402 J / g \cdot ^oC$

4 0

4 years ago