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3 years ago

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a 100gm copper block is heated in boiling water for 10min and then it is dropped into 150gm of water at 30 C in a 200gm calorime

ter.if the temperature of water is raised to 33.6 C . determine the specific heat of material of calorimeter?

1 answer:

Gekata [30.6K]3 years ago

8 0

Don't ask me son i don't know

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An ice cube at 0c was dropped into 30.0 g of water in a cup at 45.0c. at the instant that all of the ice was melted, the tempera

Ede4ka [16]

The amount of heat given by the water to the block of ice can be calculated by using
$Q=m_w C_{sw} \Delta T_w$
where
$m_w = 30 g$ is the mass of the water
$C_{sw}=4.18 J/(g ^{\circ}C)$ is the specific heat capacity of water
$\Delta T_w = 45.0^{\circ}-19.5^{\circ}C = 20.5^{\circ}C$ is the variation of temperature of the water.

Using these numbers, we find
$Q=(30 g)(4.18 J/(g^{\circ}C))(20.5^{\circ}C)=2571 J$

This is the amount of heat released by the water, but this is exactly equal to the amount of heat absorbed by the ice, used to melt it into water according to the formula:
$Q = m_i L_f$
where $m_i$ is the mass of the ice while $L_f =334 J/g$ is the specific latent heat of fusion of the ice.
Re-arranging this formula and using the heat Q that we found previously, we can calculate the mass of the ice:
$m_i = \frac{Q}{L_f}= \frac{2571 J}{334 J/g} =7.7 g$

3 0

2 years ago

A ball is kicked at an angle of 35° with the ground.a) What should be the initial velocity of the ball so that it hits a target

stiks02 [169]

Answer:

a.18.5 m/s

b.1.98 s

Explanation:

We are given that

$\theta=35^{\circ}$

a.Let $v_0$ be the initial velocity of the ball.

Distance,x=30 m

Height,h=1.8 m

$v_x=v_0cos\theta=v_0cos35$

$v_y=v_0sin\theta=v_0sin35$

$x=v_0cos\theta\times t=v_0cos35\times t$

$t=\frac{30}{v_0cos35}$

$h=v_yt-\frac{1}{2}gt^2$

Substitute the values

$1.8=v_0sin35\frac{30}{v_0cos35}-\frac{1}{2}(9.8)(\frac{30}{v_0cso35})^2$

$1.8=30tan35-\frac{6574.6}{v^2_0}$

$\frac{6574.6}{v^2_0}=21-1.8=19.2$

$v^2_0=\frac{6574.6}{19.2}$

$v_0=\sqrt{\frac{6574.6}{19.2}}=18.5 m/s$

Initial velocity of the ball=18.5 m/s

b.Substitute the value then we get

$t=\frac{30}{18.5cos35}$

t=1.98 s

Hence, the time for the ball to reach the target=1.98 s

7 0

2 years ago

a 2000 kg truck is traveling at a velocity of 30 m/s. What velocity must a 1000 kg car have in order to have the same momentum a

Molodets [167]

The car should have a velocity of 60 m/s to attain the same momentum as that of the truck of 2000 kg.

Answer:

Explanation:

Momentum is measured as the product of mass of object with the velocity attained by that object.

Momentum of 2000 kg truck = Mass × Velocity

Momentum of 2000 kg truck = 2000×30 = 60000 N

Similarly, the momentum of 1000 kg car will be 1000× velocity of the 1000 kg car.

Since, it is stated that momentum of 2000 kg truck is equal to the momentum of 1000 kg of car, then the velocity of 1000 kg of car can be determined by equating the momentum of car and truck.

Momentum of 2000 kg truck = Momentum of 1000 kg car

60000=1000×velocity of 1000 kg car

Velocity of 1000 kg car = 60000/1000=60 m/s

So, the car should have a velocity of 60 m/s to attain the same momentum as that of the truck of 2000 kg.

4 0

2 years ago

An electric motor converts electrical energy into (blank) energy

Delicious77 [7]

Answer:

Mechanical

Explanation:

5 0

2 years ago

A sample of n2 gas occupies a volume of 746 ml at stp. What volume would n2 gas occupy at 155 ◦c at a pressure of 368 torr?

musickatia [10]

Answer:

2.41 L

Explanation:

We can solve the problem by using the ideal gas equation, which can be rewritten as:

$\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}$

where we have:

$p_1 = 1.01\cdot 10^5 Pa$ (initial pressure is stp pressure)

$V_1 = 746 mL = 0.746 L = 7.46\cdot 10^{-4}m^3$ is the initial volume

$T_1 = 0^{\circ}=273 K$ is the initial temperature (stp temperature)

$p_2 = 368 torr = 4.9\cdot 10^4 Pa$ is the final pressure

$V_2 = ?$ is the final volume

$T=155^{\circ}=428 K$ is the final temperature

By substituting the numbers inside the formula and solving for V2, we find the final volume:

$V_2 = \frac{p_1 V_1 T_2}{T_1 p_2}=\frac{(1.01\cdot 10^5 Pa)(7.46\cdot 10^{-4} m^3)(428 K)}{(273 K)(4.9\cdot 10^4 Pa)}=2.41\cdot 10^{-3} m^3$

which corresponds to 2.41 L.

7 0

2 years ago