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Temka [501]
3 years ago
5

How much heat is contained in 100 kg of water at 60.0 °C?

Physics
1 answer:
attashe74 [19]3 years ago
6 0
First we need to write down heat capacity for water which is constant.
cp=4186 J/(kg*K)

The equation for Energy that we will be calculating is:

E=cp*m*T
where m is mass and T is absolute temperature (273,15 + 60 in this case). Replacing all the values in equation we get:

E = 4186*100*333,15 = 139 456 590 J
You might be interested in
How much work is done when 0.0050 c is moved through a potential difference of 9.0 v? use w = qv?
quester [9]

Answer:

0.045 J

Explanation:

The work done on a charge moving through a potential difference is given by

W=q\Delta V

where

W is the work done

q is the charge

\Delta V is the potential difference

In this problem, we have

q = 0.0050 C is the charge

\Delta V=9.0 V is the potential difference

Using the formula, we find the work done:

W=(0.0050 C)(9.0 V)=0.045 J

4 0
3 years ago
Read 2 more answers
A negative charge of -0.550 m exerts an upward <a href="/cdn-cgi/l/email-protection" class="__cf_email__" data-cfemail="5868766e
almond37 [142]

Answer:

a. +10.9μC

b. 0.600N and downward

Explanation:

To determine the magnitude of the charge, we use the force rule that exist between two charges which us expressed as

F=(kq₁q₂)/r²

since q₁=-0.55μC and the force it applied on the charge above it is upward,we can conclude that the second charge is +ve, hence we calculate its magnitude as

q₂=Fr²/kq₁

q₂=(0.6N*0.3²)/(9*10⁹*0.55*10⁻⁶)

q₂=0.054/4950

q₂=1.09*10⁻⁵c

q₂=10.9μC.

Hence the second charge is +10.9μC

b. From the rule of charges which state that like charges repel and unlike charges attract, we can conclude that the two above charges will attract since they are unlike charges. Hence the direction of the force will be downward into the second charge and the magnitude of the force will remain the same as 0.600N

8 0
3 years ago
You've always wondered about the acceleration of the elevators in the 101-story-tall Empire State Building. One day, while visit
kondor19780726 [428]

Answer:

(i) The maximum acceleration upward is 2.02 m/s².

(ii) The maximum acceleration downward is 1.39 m/s².

Explanation:

Let a be the maximum acceleration of the elevator.

The mass of the person at ground is 150 lb. We have to convert the mass into kg,

1 lb = 0.453592 kg

150 lb = 68 kg

170 lb = 77 kg

120 lb = 54 kg

(i) The person experience a force due to Earth's gravity in the downwards direction. The magnitude of this force is:

F₁ = mg = 68 x 9.8 = 666.4 N

The weight of the person decreases as the elevator is moving upwards. So, the force experienced by the person in this case due to gravity is:

F₂ = 54 x 9.8 = 529.2 N

Applying Newton's force equation;

(F₁ - F₂) = ma

(666.4 - 529.2) = 68 x a

a = 2.02 m/s²

(ii) The weight of the person increases as the elevator is moving downwards. So, the force experienced by the person in this case due to gravity is:

F₂ = 77 x 9.8 = 754.6 N

Applying Newton's force equation;

(F₁ - F₂) = ma

(666.4 - 754.6) = 68 x a

a = -1.30 m/s²

8 0
4 years ago
A 13.0 g piece of Styrofoam carries a net charge of â5.40 μC and floats above the center of a large horizontal sheet of plastic
kati45 [8]

Answer:

σ =4.180×10^{-9} C/m^2

Explanation:

electric field due to non conducting sheet is

E=\frac{\sigma}{2\epsilon_0}

the force acting on the piece of Styrofoam

Eq= mg

⇒E= mg/q

now,

\frac{mg}{q} = \frac{\sigma}{2\epsilon_0}

⇒\sigma= \frac{2mg\epsilon_0}{q}

\sigma= \frac{2\times0.013\times9.81\times8.85\times10^{-12}}{5.40\times10^{-4}}

charge per unit area (in C/m2) on the plastic sheet σ =4.180×10^{-9} C/m^2

6 0
3 years ago
A force of 120 N is exerted on a 40 kg container which sits on a floor. If the frictional force between floor and container is 8
harkovskaia [24]

Answer:

Magnitude is 144N .

Explanation:

a²+b²=c²

120N²+80N²=c²

14400+6400=c²

\sqrt{20800}   =  \sqrt{c} ^{2}

144N=c

7 0
3 years ago
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