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3 years ago

Calculate the heat required to change 2.0kg of ice at 0°C to water at 0°C

Physics

1 answer:

g100num [7]3 years ago

5 0

Answer:

160000 calories or 669440 Joules

Explanation:

Using the formula as follows;

Q = L × m

Where;

Q = amount of heat (J)

L = specific latent heat

m = mass (g)

* Latent heat (L) required for melting of ice is 80 calories/grams.

* Mass of ice in grams = 2kg × 1000 = 2000g.

Hence, Q = 80 × 2000

= 160000 Calories

1 calorie = 4.184J

160000 calorie = 160000 × 4.184

= 669440J

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how did we discover how to brake the sound barior and why did we not go as fast as the speed of light

quester [9]

The problem of the "sound barrier" has to do with moving through
air, and the things the air does when you try to push it out of the
way faster than the speed of sound. Moving through air faster
than sound was an engineering and technological problem, not
a scientific one.

Concerning light, that's about 874 thousand times faster.
See the problem ?

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3 years ago

A 5 kg fish swimming at 1 m/s swallows an absentminded 500 g fish swimming toward it at a velocity that brings both fish to a ha

AlexFokin [52]

To solve this problem we will apply the concepts related to the conservation of momentum. Momentum is defined as the product between mass and velocity of each body. And its conservation as the equality between the initial and final momentum. Mathematically described as

$m_1u_1+m_2u_2 = (m_1+m_2)v_f$

Here

$m_1$ = Mass of big fish

$m_2$ = Mass of small fish

$v_1$ = Velocity of big fish

$v_2$ = Velocity of small fish

$v_F$ = Final Velocity

The big fish eats small fish and the final velocity is zero. Rearrange the equation for the initial velocity of small fish we have

$m_1u_1=-m_2u_2$

$u_2 = -\frac{m_1u_1}{m_2}$

Replacing we have,

$u_2 = -\frac{(5kg)(1m/s)}{0.5kg}$

$u_2 = -10m/s$

The negative sign indicates that the small fish is swimming in the direction opposite to that of the big fish.

Therefore the speed of the small fish is 10m/s

8 0

3 years ago

How much work is needed to lift a 3kg create a vertical displacement of 22m

Ulleksa [173]

I think its W= 647.239J

6 0

3 years ago

Suppose someone pours 0.250 kg of 20.0ºC water (about a cup) into a 0.500-kg aluminum pan with a temperature of 150ºC. Assume th

Troyanec [42]

Answer : The temperature when the water and pan reach thermal equilibrium short time later is, $59.10^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of aluminium = $0.90J/g^oC$

$c_2$ = specific heat of water = $4.184J/g^oC$

$m_1$ = mass of aluminum = 0.500 kg = 500 g

$m_2$ = mass of water = 0.250 kg = 250 g

$T_f$ = final temperature of mixture = ?

$T_1$ = initial temperature of aluminum = $150^oC$

$T_2$ = initial temperature of water = $20^oC$

Now put all the given values in the above formula, we get:

$500g\times 0.90J/g^oC\times (T_f-150)^oC=-250g\times 4.184J/g^oC\times (T_f-20)^oC$

$T_f=59.10^oC$

Therefore, the temperature when the water and pan reach thermal equilibrium short time later is, $59.10^oC$

8 0

4 years ago

A 80 W light bulb (normally run at 120 V) is attached to a transformer. The voltage source in the transformer is 65 V and Np = 3

Marina CMI [18]

67.8 turns needed by the secondary coil to run the bulb.

<u>Explanation</u>:

We know that,

$\text { Electric power }(p)=\frac{V^{2}}{R}$

$\text { Hence, } \frac{P_{1}}{P_{2}}=\frac{V_{1}^{2} / R}{V_{2}^{2} / R}$

$\frac{P_{1}}{P_{2}}=\frac{V_{1}^{2}}{V_{2}^{2}}$

For calculating number of turns

$\frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}$

Given that,

$80 \mathrm{W}\left(P_{1}\right) \text { bulb with voltage } 120 \mathrm{V}\left(V_{1}\right) \text { is connected to a transformer. }$

$\text { The source voltage of a transformer is }\left(V_{P}\right) \text { is } 65 \mathrm{V}$

$\text { The number of turns in primary winding of transformer is }\left(N_{P}\right) \text { is } 30 .$

We need to find the number of turns in the secondary winding $\left(N_{S}\right)$ to run the bulb at 120W $\left(P_{2}\right)$

Firstly find the secondary voltage in the transformer use, $\frac{P_{1}}{P_{2}}=\frac{V_{1}^{2}}{V_{2}^{2}}$

$\frac{80}{120}=\frac{120^{2}}{V_{2}^{2}}$

$V_{2}^{2}=\frac{120^{2} \times 120}{80}$

$V_{2}^{2}=\frac{1728000}{80}$

$V_{2}^{2}=21600$

$V_{2}=\sqrt{21600}$

$V_{2}=146.9 \mathrm{V}=V_{S}$

Now, finding the number of turns in secondary coil. Use, $\frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}$

$\frac{30}{N_{S}}=\frac{65}{146.9}$

$N_{S}=\frac{30 \times 146.9}{65}$

$N_{S}=\frac{4407}{65}$ $N_{S}=67.8$

The number of turns in the secondary winding are 67.8 turns.

6 0

3 years ago