The heat energy required to convert 10 ,kg of | vedclass

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

Question

(A) Mass of ice,$m = 10 \,kg = 10^4 \,g$.Specific heat capacity of ice,$c = 0.5 \,cal \,g^{-1} \,^{\circ}C^{-1} = 2093.4 \,J \,kg^{-1} \,K^{-1}$.Laten

Vedclass · Accepted Answer

$357 	imes 10^4 \,J$

Vedclass · Answer

(A) Mass of ice,$m = 10 \,kg = 10^4 \,g$.Specific heat capacity of ice,$c = 0.5 \,cal \,g^{-1} \,^{\circ}C^{-1} = 2093.4 \,J \,kg^{-1} \,K^{-1}$.Latent heat of fusion of ice,$L = 80 \,cal \,g^{-1} = 334944 \,J \,kg^{-1}$.The total heat energy $Q$ required is the sum of heat to raise the temperature of ice to $0^{\circ}C$ and the heat to melt the ice at $0^{\circ}C$.$Q = m c \Delta T + m L$$Q = 10 \,kg 	imes 2093.4 \,J \,kg^{-1} \,K^{-1} 	imes (0 - (-10)) \,K + 10 \,kg 	imes 334944 \,J \,kg^{-1}$$Q = 10 	imes 2093.4 	imes 10 + 3349440$$Q = 209340 + 3349440$$Q = 3558780 \,J \approx 357 	imes 10^4 \,J$.

The heat energy required to convert $10 \,kg$ of ice at $-10^{\circ} C$ into water at $0^{\circ} C$ is (specific heat capacity of ice $=0.5 \,cal \,g^{-1} \,^{\circ}C^{-1}$ and latent heat of fusion of ice $=80 \,cal \,g^{-1}$)

Similar Questions

Expansion during heating:

In cold countries,water pipes sometimes burst,because

The pressure applied from all directions on a cube is $P$. How much should its temperature be raised to maintain the original volume? The volume elasticity of the cube is $\beta$ and the coefficient of volume expansion is $\alpha$.

It is known that wax contracts on solidification. If molten wax is taken in a large vessel and it is allowed to cool slowly,then

When $1\, g$ of water at $0^{\circ}C$ and $1 \times 10^5\, N/m^2$ pressure is converted into ice of volume $1.091\, cm^3$,the external work done will be (in $, J$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module