1. Tardigrade
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  4. How much thermal energy is required to change a 40 g ice cube from solid at -10° C to steam at 110° C. [Assume latent heat of fusion for water =80 kcal / kg, specific heat of water =1 k cal / kg ° C, specific heat of Ice =0.5 k cal kg ° C, specific heat of steam =0.48 k cal / kg ° C, latent heat of vaporisation of water- 540 kcal / kg ° C ]

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Q. How much thermal energy is required to change a $40\, g$ ice cube from solid at $-10^{\circ} C$ to steam at $110^{\circ} C$.
[Assume latent heat of fusion for water $=80 \,kcal / kg$, specific heat of water $=1\, k\,cal / kg { }^{\circ} C$,
specific heat of Ice $=0.5\, k\,cal kg { }^{\circ} C$,
specific heat of steam $=0.48\, k\,cal / kg ^{\circ} C$,
latent heat of vaporisation of water- $540 \,kcal / kg { }^{\circ} C$ ]

TS EAMCET 2020

Solution:

Following changes occur in given conversion process
image
So, required thermal energy
$=m\left\{c _ { \text { ice } } \left(0^{\circ} C -\left(-10^{\circ} Q \right)+\right.\right.$
$L_{\text {fusion }}+c_{\text {water }}\left(100^{\circ} C -0^{\circ} C \right) +L_{\text {vaporisation }}\left.+c_{\text {stcam }}\left(110^{\circ} C -100^{\circ} C \right)\right\} $
$=\frac{400}{1000}\{0.5 \times 10+80+1 \times 100+540+0.48 \times 10\} $
$=29.192\, k\,cal$