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

Byju's Answer

Standard X

Physics

Calorimeter

A small quant...

Question

A small quantity, mass $m$ , of water at a temperature $θ (i n^{\circ} C)$ is poured on to a large mass $M$ of ice which is at its melting point. If $c$ is specific heat capacity of water and $L$ the latent heat of fusion of ice, then the mass of ice melted is given by

\frac{M L}{m c θ}

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

\frac{m c θ}{M L}

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

\frac{M c θ}{L}

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

\frac{m c θ}{L}

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

Open in App

Solution

The correct option is C $\frac{m c θ}{L}$ .
Heat gain is $m c (θ - 0)$
Heat loss if ice is $m_{1} L; m_{1}$ is ice melted.
$m_{1} L = m c θ$
$\Rightarrow m_{1} = m c θ / L$

Suggest Corrections

Similar questions

Q. 80

g

of water at

30^{\circ} C

is poured on a large block of ice at

0^{\circ} C

. The mass of ice that melts is (Latent heat of fusion of ice is

80 c a l / g

and specific heat of water is

1 c a l / g^{\circ} C

)

80 g

of water at

30^{o} C

is poured on large block of ice at

0^{\circ}

C. The mass of ice that melts is

Given, Latent heat =

334 J / g

Specific heat of water

= 4.2 J g^{- 1} C^{o - 1}

Q. A piece of ice of mass

40 g

0^{o} C

is added to

200 g

of water at

50^{\circ} C

. Calculate the final temperature of water when all the ice has melted. Specific heat capacity of water =

4200 J k g^{- 1} K^{- 1}

and specific latent heat of fusion of ice =

336 \times 10^{3} J k g^{- 1}

2 k g

of ice melts when water at

100^{\circ} C

is poured in a hole drilled in a block of ice. What mass of water was used?

Given: specific heat capacity of water =

4200 J k g^{- 1} K^{- 1}

specific latent heat of ice =

336 \times 10^{3} J k g^{- 1}

A piece of ice of mass 40 g is dropped into 200 g of water at 50 $^{\circ} C$ . Calculate the final temperature of the water after all the ice has melted. (Specific heat capacity of water = 4200 J/kg $^{\circ} C$ , specific latent heat of fusion of ice $= 336 \times 10^{3}$ J/kg)

Join BYJU'S Learning Program

A small quantity, mass m, of water at a temperature θ(in∘C) is poured on to a large mass M of ice which is at its melting point. If c is specific heat capacity of water and L the latent heat of fusion of ice, then the mass of ice melted is given by

The correct option is C mcθL.Heat gain is mc(θ−0)Heat loss if ice is m1L;m1 is ice melted.m1L=mcθ⇒m1=mcθ/L

A small quantity, mass $m$ , of water at a temperature $θ (i n^{\circ} C)$ is poured on to a large mass $M$ of ice which is at its melting point. If $c$ is specific heat capacity of water and $L$ the latent heat of fusion of ice, then the mass of ice melted is given by

The correct option is C $\frac{m c θ}{L}$ .
Heat gain is $m c (θ - 0)$
Heat loss if ice is $m_{1} L; m_{1}$ is ice melted.
$m_{1} L = m c θ$
$\Rightarrow m_{1} = m c θ / L$