Advertisements
Advertisements
Question
500 g of molten lead at its m.p. is just kept in molten state by a heater of 25 W. If the heater is switched off, the temperature of lead starts falling after 7 min. Calculate the sp. latent heat of fusion of lead.
Numerical
Advertisements
Solution
Time for which temp. does not fall = Time in which lead changes its state from liquid to solid.
Heat given out by lead at its m.p. = mL
= 500 g × L
Also, heat given out by lead = P × t
\[=25 \frac{J}{s} \times 7 \times 60 \mathrm{~s}\]
= 10500 J
∴ 500 g × L = 10500 J
\[\therefore L =\frac{10500}{500}\]
= 21 Jg−1
shaalaa.com
Is there an error in this question or solution?
