Question

a piece of lead of mass 120 gm is given 1000 calorie of energy.calculate the rise is it's temperature.the value of lead=0.13×1000jkg-1k-1​

Answer 1

Answer:

The rise in temperature of the lead is given as 269.2 degree C

Explanation:

As we know that the heat required to raise the temperature of the given matter is given by

$Q = ms\Delta T$

now we know that

m = 120 g

$s = 130 J/kg K$

now we have

$Q = \frac{120}{1000}(130)\Delta T$

$Q = 15.6 \Delta T$

so above is the heat required to raise the temperature by T

$Q = 1000 Cal = 4200 J$

$4200 = 15.6 \Delta T$

$\Delta T = 269.2 ^oC$

#Learn

Topic : Heat calculation

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