A block of mass 2.5,, kg is heated to tempera | vedclass

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Question

Here, mass of copper block,

$M=2.5 \mathrm{kg}=2.5 	imes 10^{3} \mathrm{g}$

Rise in temperature of the copper block,

$\Delta T=500^{\circ} \

Vedclass · Accepted Answer

$1.5$

Vedclass · Answer

Here, mass of copper block,

$M=2.5 \mathrm{kg}=2.5 	imes 10^{3} \mathrm{g}$

Rise in temperature of the copper block,

$\Delta T=500^{\circ} \mathrm{C}-0^{\circ} \mathrm{C}=500^{\circ} \mathrm{C}$

Specific heat of copper, $c=0.39 J / g^{\circ} C$

Lalent heat of fusion of water, $L=335 J / g$

if $m$ is the mass of ice melted, applying the principle of calorimetry.

Heat gained by ice=heat lost by copper block

i.e., $m L=M c \Delta T$

or, $m=\frac{M c \Delta T}{L}$

$=\frac{2.5 	imes 10^{3}(g) 	imes 0.39\left(J / g C^{0}ight) 	imes 500\left(C^{\circ}ight)}{335(J / g)}$

$=1455 g=1.455 \approx 1.5 k g$

A block of mass $2.5\,\, kg$ is heated to temperature of $500^o C$ and placed on a large ice block. ......... $kg$ is the maximum amount of ice that can melt (approx.). Specific heat for the body $= 0.1 Cal/gm^o C$.

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