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Cerrena [4.2K]
3 years ago
11

Suppose 8.50 ✕ 10^5 J of energy are transferred to 1.63 kg of ice at 0°C. The latent heat of fusion and specific heat of water a

re Lf = 3.33 ✕ 105 J/kg and c = 4186 J (kg · °C) . HINT (a) Calculate the energy (in J) required to melt all the ice into liquid water. (Enter your answer to at least three significant figures.) J (b) How much energy (in J) remains to raise the temperature of the liquid water? (Enter your answer to at least three significant figures.) J (c) Determine the final temperature of the liquid water in Celsius. °C
Physics
1 answer:
PolarNik [594]3 years ago
5 0

Answer:

(a) 5.43 x 10⁵ J

(b) 3.07 x 10⁵ J

(c) 45 °C

Explanation:

(a)

L_{f} = Latent heat of fusion of ice to water = 3.33 x 10⁵ J/kg

m = mass of ice = 1.63 kg

Q_{f} = Energy required to melt the ice

Energy required to melt the ice is given as

Q_{f} = m L_{f}

Q_{f} = (1.63) (3.33 x 10⁵)

Q_{f} = 5.43 x 10⁵ J

(b)

E = Total energy transferred = 8.50 x 10⁵ J

Q  = Amount of energy remaining to raise the temperature

Using conservation of energy

E = Q_{f} + Q

8.50 x 10⁵ = 5.43 x 10⁵ + Q

Q = 3.07 x 10⁵ J

(c)

T₀ = initial temperature = 0°C

T = Final temperature

m = mass of water = 1.63 kg

c = specific heat of water = 4186 J/(kg °C)

Q = Amount of energy to raise the temperature of water = 3.07 x 10⁵ J

Using the equation

Q = m c (T - T₀)

3.07 x 10⁵ = (1.63) (4186) (T - 0)

T = 45 °C

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Running at 1.55 m/s, Bruce, the 40.0 kg quarterback, collides with Biff, the 90.0 kg tackle, who is traveling at 7.0 m/s in the
Margarita [4]

Answer:Bruce is knocked backwards at  

14

m

s

.

Explanation:

This is a problem of momentum (

→

p

) conservation, where

→

p

=

m

→

v

and because momentum is always conserved, in a collision:

→

p

f

=

→

p

i

We are given that  

m

1

=

45

k

g

,  

v

1

=

2

m

s

,  

m

2

=

90

k

g

, and  

v

2

=

7

m

s

The momentum of Bruce (

m

1

) before the collision is given by

→

p

1

=

m

1

v

1

→

p

1

=

(

45

k

g

)

(

2

m

s

)

→

p

1

=

90

k

g

m

s

Similarly, the momentum of Biff (

m

2

) before the collision is given by

→

p

2

=

(

90

k

g

)

(

7

m

s

)

=

630

k

g

m

s

The total linear momentum before the collision is the sum of the momentums of each of the football players.

→

P

=

→

p

t

o

t

=

∑

→

p

→

P

i

=

→

p

1

+

→

p

2

→

P

i

=

90

k

g

m

s

+

630

k

g

m

s

=

720

k

g

m

s

Because momentum is conserved, we know that given a momentum of  

720

k

g

m

s

before the collision, the momentum after the collision will also be  

720

k

g

m

s

. We are given the final velocity of Biff (

v

2

=

1

m

s

) and asked to find the final velocity of Bruce.

→

P

f

=

→

p

1

f

+

→

p

2

f

→

P

f

=

m

1

v

1

f

+

m

2

v

2

f

Solve for  

v

1

:

v

1

f

=

→

P

f

−

m

2

v

2

f

m

1

Using our known values:

v

1

f

=

720

k

g

m

s

−

(

90

k

g

)

(

1

m

s

)

45

k

g

v

1

f

=

14

m

s

∴

Bruce is knocked backwards at  

14

m

s

.

Explanation:

5 0
3 years ago
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