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3 years ago

7

A copper pot with a mass of 0.475 kg contains 0.185 kg of water, and both are at a temperature of 19.0 ∘C . A 0.240 kg block of

iron at 86.5 ∘C is dropped into the pot.Find the final temperature of the system, assuming no heat loss to the surroundings.

1 answer:

____ [38]3 years ago

7 0

Answer:

The final temperature is 25.85°C.

Explanation:

Given that,

Mass of copper pot = 0.475 kg

Mass of water = 0.185 kg

Temperature = 19.0°C

Mass of iron block = 0.240 kg

Temperature = 86.5°C

We need to calculate the final temperature

Using formula of heat

Heat lost by iron block=Heat gained by copper pot and water

$-M_{i}C_{i}\Delta T=M_{w}C_{w}\Delta T+M_{c}C_{c}\Delta T$

Put the value into the formula

$-0.240\times450\times(T-86.5)=0.185\times4180\times(T-19.0)+0.475\times384\times(T-19.0)$

$-108\timesT+108\times86.5=773.3\times T-773.3\times19.0+182.4\times T-182.4\times19.0$

$9342-108T=773.3T-14692.7+182.4T-3465.6$

$-108T-773.3T-182.4T=-9342-14692.7-3465.6$

$1063.7T=27500.3$

$T=\dfrac{27500.3}{1063.7}$

$T=25.85^{\circ}C$

Hence, The final temperature is 25.85°C.

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Two particles with masses 2m and 9m are moving toward each other along the x axis with the same initial speeds vi. Particle 2m i

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Answer:

The final speed for the mass 2m is $v_{2y}=-1,51\ v_{i}$ and the final speed for the mass 9m is $v_{1f} =0,85\ v_{i}$ .

The angle at which the particle 9m is scattered is $\theta = -66,68^{o}$ with respect to the - y axis.

Explanation:

In an elastic collision the total linear momentum and the total kinetic energy is conserved.

<u>Conservation of linear momentum:</u>

Because the linear momentum is a vector quantity we consider the conservation of the components of momentum in the x and y axis.

The subindex 1 will refer to the particle 9m and the subindex 2 will refer to the particle 2m

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$p_{yf} =9m\ v_{1y} - 2m\ v_{2y}$

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$p_{yi} =p_{yf} \\0=9m\ v_{1y} -2m\ v_{2y} \\v_{1y}=\frac{2}{9} \ v_{2y}$

<u>Conservation of kinetic energy:</u>

$\frac{1}{2}\ 9m\ v_{i} ^{2} +\frac{1}{2}\ 2m\ v_{i} ^{2}=\frac{1}{2}\ 9m\ v_{1f} ^{2} +\frac{1}{2}\ 2m\ v_{2f} ^{2}$

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