At a power station,heat is removed from the h | vedclass

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

Question

(B) The rate of heat removal is given by $Q/t = 6.7 	imes 10^9 \ J/min$.To convert this to $J/s$,we divide by $60$: $\frac{dQ}{dt} = \frac{6.7 	imes

Vedclass · Accepted Answer

$\frac{6.7 	imes 10^9}{4200 	imes 8 	imes 60} \ kg \ s^{-1}$

Vedclass · Answer

(B) The rate of heat removal is given by $Q/t = 6.7 	imes 10^9 \ J/min$.To convert this to $J/s$,we divide by $60$: $\frac{dQ}{dt} = \frac{6.7 	imes 10^9}{60} \ J/s$.The formula for heat transfer is $\frac{dQ}{dt} = \frac{dm}{dt} 	imes S 	imes \Delta T$,where $\frac{dm}{dt}$ is the rate of water flow.Given $S = 4200 \ J/kg \ ^oC$ and $\Delta T = 14.0 \ ^oC - 6.0 \ ^oC = 8.0 \ ^oC$.Substituting the values: $\frac{6.7 	imes 10^9}{60} = \frac{dm}{dt} 	imes 4200 	imes 8$.Rearranging for the rate of water flow: $\frac{dm}{dt} = \frac{6.7 	imes 10^9}{4200 	imes 8 	imes 60} \ kg \ s^{-1}$.

At a power station,heat is removed from the heat exchanger by cooling water at $6.7 \times 10^9 \ J$ per minute. The cooling water enters at $6.0 \ ^oC$ and leaves at $14.0 \ ^oC$. [Take the specific heat capacity of water $4200 \ J/kg \ ^oC$]. Which of the following is the rate of water flow?

Similar Questions

$50\, g$ 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 ........ $g$.

$2 \; g$ of steam condenses when passed through $40 \; g$ of water initially at $25 \; ^\circ C$. The condensation of steam raises the temperature of water to $54.3 \; ^\circ C$. What is the latent heat of steam in $cal/g$?

The $SI$ unit of mechanical equivalent of heat is

What is calorimetry? What is a calorimeter? Explain its principle and construction.

Vedclass Test Series

Exam Paper Generator

Online Exam Module