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En un experimento de calorimetría, 0.50 kg de un metal a 100°C se añaden a 0.50 kg de agua a 20°C en un vaso de calorímetro de a

Maru [420]

Answer:

c=0.14J/gC

Explanation:

A.

2) The specific heat will be the same because it is a property of the substance and does not depend on the medium.

B.

We can use the expression for heat transmission

$Q=mc(T_2-T_1)$

In this case the heat given by the metal (which is at a higher temperature) is equal to that gained by the water, that is to say

$Q_1=-Q_2$

for water we have to

c = 4.18J / g ° C

replacing we have

$c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}$

I hope this is useful for you

A.

2) El calor específico será igual porque es una propiedad de la sustancia y no depende del medio.

B.

Podemos usar la expresión para la transmisión de calor

$Q=mc(T_2-T_1)$

En este caso el calor cedido por el metal (que está a mayor temperatura) es igual al ganado por el agua, es decir

$Q_1=-Q_2$

para el agua tenemos que

c=4.18J/g°C

reemplazando tenemos

$c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}$

7 0

3 years ago

Bart stole a watermelon and ran 5,000 feet from the cops and they chase lasted 0.1 hours how fast was Bart running in miles per

liraira [26]

I am not as sure but I think it is 9.469 miles

5 0

3 years ago

A stone is dropped from the edge of a roof, and hits the ground with a velocity of -180 feet per second. How high (in feet) is t

Neko [114]

Answer:

d = 506.25 ft

Explanation:

As we know by kinematics that

$v_f^2 - v_i^2 = 2 a d$

here we know that initially the stone is dropped from rest from the edge of the roof

so here initial speed will be zero

now we have

$v_i = 0$

also the acceleration of the stone is due to gravity which is given as

$g = 32 ft/s^2$

now we have

$v_f = 180 ft/s$

so from above equation

$180^2 - 0 = 2(32)d$

$d = 506.25 ft$

6 0

3 years ago

Qual a capacidade térmica de um objeto que, ao receber 10000 cal de energia, tem sua temperatura elevada de 25°C para 75°C?

Stolb23 [73]

Answer:

$200 cal/^{\circ}C$

Explanation:

When heat energy is supplied to an object, the temperature of the object increases according to the equation:

$Q=C\Delta T$

where

Q is the heat supplied

C is the heat capacity of the object

$\Delta T$ is the change in temperature

In this problem we have:

$Q=10,000 cal$ is the energy supplied

$\Delta T=75C-25C=50C$ is the change in temperature of the object

Therefore, the heat capacity of the object is:

$C=\frac{Q}{\Delta T}=\frac{10,000}{50}=200 cal/^{\circ}C$

6 0

3 years ago

(c) Due to up thrust

garri49 [273]

Answer:

B1. Pascal's law is a principal in fluid mechanics given by Blaise Pascal that states that a pressure change at any point in a confined incompressible fluid is transmitted throughout the fluid that the same change occur everywhere. 2 applications of Pascal's law are hydraulic lifts, hydraulic jacks, hydraulic hydraulic brakes ,hydraulic pumps. mark me as a braintalist list plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz

5 0

3 years ago