At 3 0 -C- a lead bullet of 50 g- is fired vertically upwards with a speed of 840 m-s- The specific heat of lead is 0-02 cal- g -C- on returning to the starting level- it strikes to a cake of ice at 0-C- Calculate the amount of ice melted Assume all the energy is spent in melting only

The correct option is D 52.875 g Kinetic energy of bullet = mv^22 = 12×0.02×840^2 = 17640 joule =4200 cal heat supplied by lead to ice : ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard VII

Physics

Heat

At 3 0 ∘C, ...

Question

At $3 0^{\circ} C$ , a lead bullet of $50 g$ , is fired vertically upwards with a speed of $840 m / s$ . The specific heat of lead is $0.02 c a l / g^{\circ} C$ . on returning to the starting level, it strikes to a cake of ice at $0^{\circ} C$ . Calculate the amount of ice melted (Assume all the energy is spent in melting only)

62.7 g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

55 k g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

52.875 k g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

52.875 g

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $52.875 g$
Kinetic energy of bullet = $\frac{m v^{2}}{2}$ = $\frac{1}{2} \times 0.02 \times 840^{2}$ = $17640 j o u l e = 4200 c a l$

heat supplied by lead to ice : $Δ H_{1} = m s Δ T$ = $50 \times 0.02 \times 30$ = $30 c a l$

Total heat supplied = $4200 + 30 = 4230 c a l$

let $M_{i}$ mass of ice melted $L$ is the latent heat of ice

$M_{i} L = 4230$
$M_{i} \times 80 = 4230$

$M_{i} = 52.875 g$

Suggest Corrections

Similar questions

Q. A bullet of mass

50 g

and specific heat capacity

800 {J kg}^{- 1} K^{- 1}

is initially at a temperature

20^{\circ} C

. It is fired vertically upwards with a speed of

200 m/s

and on returning to the starting point strikes a lump of ice at

0^{\circ} C

and gets embedded in it. Assume that all the energy of the bullet is used up in melting. Neglect the friction of air. Latent heat of fusion of ice

= 3.36 \times 10^{5} {J kg}^{- 1}

Q. A bullet of mass

50 g

and specific heat capacity

800 {J kg}^{- 1} K^{- 1}

is initially at a temperature

20^{\circ} C

. It is fired vertically upwards with a speed of

200 m/s

and on returning to the starting point strikes a lump of ice at

0^{\circ} C

and gets embedded in it. Assume that all the energy of the bullet is used up in melting. Neglect the friction of air. Latent heat of fusion of ice

= 3.36 \times 10^{5} {J kg}^{- 1}

Q. A bullet of 50 g is fired from a revolver with a velocity of 100

m s^{- 1}

into an ice block at

0^{\circ} C

. The bullet stops after it gets embedded into ice. Assuming that all the kinetic energy of the bullet is converted into heat energy which is absorbed by the melting ice, find the quantity of the ice that just melted.

Q. A lead bullet at

100^{o} C

strikes a block of wood, gets embedded in it and melts. Find the speed of the bullet if 50% of the heat produced is shared by the bullet. Specific heat of lead

= 0.03 c a l / g^{o} C

, latent heat of lead

= 5 c a l / g

and melting point of lead

= 327^{o} C

A lead bullet penetrates into a solid object and melts. Assuming that 50% of its kinetic energy was used to heat it, calculate the initial speed of the bullet. The initial temperature of the bullet is $27^{\circ} C$ and its melting point is $327^{\circ} C$ . Latent heat of fusion of lead = $2.5 \times 10^{4} J k g^{- 1}$ and specific heat capacity of lead = $125 J k g^{1} K^{1}$ .

Join BYJU'S Learning Program

At 30∘C, a lead bullet of 50 g, is fired vertically upwards with a speed of 840 m/s. The specific heat of lead is 0.02 cal/g∘C. on returning to the starting level, it strikes to a cake of ice at 0∘C. Calculate the amount of ice melted (Assume all the energy is spent in melting only)

The correct option is D 52.875 gKinetic energy of bullet = mv22=12×0.02×8402 = 17640joule=4200calheat supplied by lead to ice : ΔH1=msΔT = 50×0.02×30 = 30calTotal heat supplied = 4200+30=4230callet Mi mass of ice melted L is the latent heat of iceMiL=4230Mi×80=4230Mi=52.875g