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Burns produced by steam at 100°C are much more severe than those produced by the same mass of 100°C water. Calculate the quantity of heat in (Cal or kcal) that must be removed from 6.1 g of 100°C steam to condense it and lower its temperature to 46°C. Specific heat of water = 1.00 kcal/(kg · °C); heat of vaporization = 539 kcal/kg; specific heat of human flesh = 0.83 kcal/(kg · °C).

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Answer:

Quantity of heat is 329.4 Kcal .

Explanation:

To "calculate the amount of heat" released in a "chemical reaction", use the equation , Q = mc ΔT

Where, "Q" is the "heat energy" transferred (in joules),  

"m" is the "mass of the liquid" being heated (in grams),  

"c" is the "specific heat capacity" of the liquid (joule per gram degrees Celsius)  

"ΔT" is the change in "temperature "

Q = mc([tex]T_2[/tex] - [tex]T_1[/tex] )

Given values

m = 6.1 gram

[tex]\begin{array}{l}{\left.c=1.00 \mathrm{kcal} / \mathrm{kg} .^{\circ} \mathrm{C}\right)} \\ {\mathrm{T}_{2}=100^{\circ} \mathrm{C}} \\ {\mathrm{T}_{1}=46^{\circ} \mathrm{C}}\end{array}[/tex]

Q = 6.1 × 1.00(100 - 46)

Q = 6.1 × 54

Q = 329.4 Kcal

Quantity of heat is 329.4 Kcal .

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