To solve this problem we will apply the concept related to the heat transferred to a body to reach a certain temperature. This concept is shaped by the energy ratio of a body which is the product of the mass, its specific heat and the change in temperature. For the specific case, it will be the sum of the heat transferred to the Water, the Aluminum and the loss due to latency due to vaporization in the water. That is to say,

$\Delta Q = m_{Al} C_p \Delta T +m_wC_w \Delta T +m_w L_v$

Here,

$m_{Al}$ = Mass of Aluminum

$C_p$ = Specific Heat of Aluminum

$C_w$ = Specific Heat of Water

$m_w =$ Mass of water

$L_v =$ Latent of Vaporization

Replacing,

$\Delta Q = (0.85)(900)(100-45)+(2)(2000)(100-45)+(0.7)(2258000)$

Converting,

$\Delta Q = 1842675J (\frac{0.000239006kCal}{1J})$

$\Delta Q = 440.409kCal$

Therefore is required 440.409kCal

8 0

3 years ago

The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.420 with the floor. If

bonufazy [111]

Answer:

The distance is $s= 30.3 \ m$

Explanation:

From the question we are told that

The coefficient of static friction is $\mu_s = 0.42$

The initial speed of the train is $u = 57 \ km /hr = 15.8 \ m/s$

For the crate not to slide the friction force must be equal to the force acting on the train i.e

$-F_f = F$

The negative sign shows that the two forces are acting in opposite direction

=> $mg * \mu_s = ma$

=> $-g * \mu_s = a$

=> $a = -9.8 * 0.420$

=> $a = -4.116 m/s^2$

From equation of motion

$v^2 = u^2 + 2as$

Here v = 0 m/s since it came to a stop

=> $s= \frac{v^2 - u^2 }{ 2 a}$

=> $s= \frac{0 -(15.8)^2 }{ - 2 * 4.116}$

=> $s= 30.3 \ m$

7 0

2 years ago

In a closed system, the loss of momentum of one object ________ the gain in momentum of another object.

Virty [35]

In a closed system, the loss of momentum of one object is same as________ the gain in momentum of another object

according to law of conservation of momentum, total momentum before and after collision in a closed system in absence of any net external force, remains conserved . that is

total momentum before collision = total momentum after collision

P₁ + P₂ = P'₁ + P'₂

where P₁ and P₂ are momentum before collision for object 1 and object 2 respectively.

P'₁ - P₁ = - (P'₂ - P₂)

so clearly gain in momentum of one object is same as the loss of momentum of other object

8 0

2 years ago