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

Principle of Calorimetry

Find the amou...

Question

Find the amount of heat supplied to decrease the volume of an ice water mixture by $1 c m^{3}$ without any change in temperature. $(ρ_{i c e} = 0.9 ρ_{w a t e r}, L_{i c e} = 80 c a l / g m)$

360 c a l

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

500 c a l

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

720 c a l

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

none of these

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

Open in App

Solution

The correct option is C $720 c a l$
Since density of ice is=0.9 x density of water.
hence to decrease the the volume of ice-water mixture by $1 c m^{3}$ , $10 c m^{3}$ of ice should melt.
Mass of $10 c m^{3}$ ice is $9 g m$
heat required= $Q = m \times L = 720 c a l$ .

Suggest Corrections

Similar questions

1 {cm}^{3}

ρ_{ice} = 0.9 ρ_{water}, L_{ice} = 80 cal/gm)

1 {cm}^{3}

(Take ρ_{ice} = 0.9 ρ_{water}, L_{f} = 80 {cal g}^{- 1})

500 g

L_{i c e} = 3.34 \times 10^{5} J {k g}^{- 1}

5 g m

30^{\circ} C

5 g m

- 20^{\circ} C

= 0.5 c a l / g m^{\circ} C

= 80 c a l / g m

10

45

28^{o} C

- 15^{o} C

^{o} C

i c e = 0.5 c a l / g m /^{o} C

L = 80 c a l / g m

Join BYJU'S Learning Program