Heat required to melt 1, g of ice is 80, cal. | vedclass

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

Question

(B) The heat required to melt $60\, g$ of ice is $Q = m 	imes L_f = 60\, g 	imes 80\, cal/g = 4800\, cal$.Converting this heat into Joules, we use t

Vedclass · Accepted Answer

$336$

Vedclass · Answer

(B) The heat required to melt $60\, g$ of ice is $Q = m 	imes L_f = 60\, g 	imes 80\, cal/g = 4800\, cal$.Converting this heat into Joules, we use the conversion factor $1\, cal = 4.2\, J$:$Q = 4800 	imes 4.2\, J = 20160\, J$.The time taken is $t = 1\, 	ext{minute} = 60\, s$.Power is defined as the rate of doing work or energy consumption: $P = \frac{Q}{t}$.$P = \frac{20160\, J}{60\, s} = 336\, W$.

Heat required to melt $1\, g$ of ice is $80\, cal$. $A$ man melts $60\, g$ of ice by chewing in one minute. His power is ........ $W$.

Similar Questions

$\frac{(1+i)^{2011}}{(1-i)^{2009}}$ is equal to

Which of the following compounds is an optically inactive compound?

If $\frac{3x^2 + 5}{(x^2 + 1)^2} = \frac{a}{x^2 + 1} + \frac{b}{(x^2 + 1)^2}$,then $(a, b) = $

On what basis are the three types of lichens classified?

$A$ transistor is used as an amplifier in $CB$ mode with a load resistance of $5 \; k\Omega$. The current gain of the amplifier is $0.98$ and the input resistance is $70 \; \Omega$. The voltage gain and power gain respectively are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module