50 , g of ice at 0^{circ} C is dropped into a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Heat lost by $100 \, g$ of water to cool down from $30^{\circ} C$ to $0^{\circ} C$ is given by $Q = mc\Delta T$.Here,$m = 100 \, g$,$c = 1 \, cal/

Vedclass · Accepted Answer

$12.5$

Vedclass · Answer

(A) Heat lost by $100 \, g$ of water to cool down from $30^{\circ} C$ to $0^{\circ} C$ is given by $Q = mc\Delta T$.Here,$m = 100 \, g$,$c = 1 \, cal/g^{\circ} C$,and $\Delta T = 30^{\circ} C$.$Q = 100 	imes 1 	imes 30 = 3000 \, cal$.This heat is used to melt the ice at $0^{\circ} C$. The latent heat of fusion of ice is $L_f = 80 \, cal/g$.The mass of ice melted is $m_{melted} = \frac{Q}{L_f} = \frac{3000}{80} = 37.5 \, g$.The initial mass of ice was $50 \, g$.Therefore,the amount of ice left is $m_{left} = 50 - 37.5 = 12.5 \, g$.

$50 \, g$ of ice at $0^{\circ} C$ is dropped into a calorimeter containing $100 \, g$ of water at $30^{\circ} C$. If the thermal capacity of the calorimeter is zero,then the amount of ice left in the mixture at equilibrium is .......... $g$.

Similar Questions

The motion of a particle represented by $y = \sin \omega t - \cos \omega t$ is

If $\alpha = \frac{5}{2! 3} + \frac{5 \cdot 7}{3! 3^2} + \frac{5 \cdot 7 \cdot 9}{4! 3^3} + \ldots$,then $\alpha^2 + 4\alpha$ is equal to

The sulphate of a metal $M$ contains $9.87 \%$ of $M$. This sulphate is isomorphous with $ZnSO_4 \cdot 7 H_2 O$. The atomic weight of $M$ is (in $.3$)

Maximum bond angle is present in

The increasing order of the atomic radius of $Si, S, Na, Mg, Al$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module