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Pie
3 years ago
10

How many kilojoules of heat energy are absorbed when 98.5 g of water are heated from 24.5 oC to 48.8 oC?

Chemistry
1 answer:
Furkat [3]3 years ago
4 0

Answer:

10.01461 kilojoules of thermal energy are absorbed when 98.5 g of water is heated from 24.5 ° C to 48.8 ° C

Explanation:

Sensible heat is the amount of heat that a body absorbs or releases without any changes in its physical state (phase change) causing a change in temperature.

The equation that allows to calculate heat exchanges in this case is:

Q = c * m * ΔT

where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the change in temperature.

In this case:

  • c=4.184 \frac{J}{g*C}
  • m= 98.5 g
  • ΔT= Tfinal - Tinitial= 48.8°C - 24.5°C= 24.3 °C

Replacing:

Q= 4.184 \frac{J}{g*C} *98.5 g* 24.3 °C

Solving:

Q=10,014.61 J

Since 1 J is equal to 0.001 kJ, then you can apply the following rule of three: if 1 J is equal to 0.001 kJ, then 10,014.61 J is equal to how many kJ?

kJ=\frac{10,014.71 J*0.001 kJ}{1 J}

kJ=10.01461 kJ

<u><em>10.01461 kilojoules of thermal energy are absorbed when 98.5 g of water is heated from 24.5 ° C to 48.8 ° C</em></u>

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The explanation of the processes in which pigments are involved (capturing light and forming ATP and NADPH) is given in the following paragraphs)

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If a gas occupies 1532.7 mL at standard temperature, what volume does it occupy at 49.4 ºC if the pressure remains constant?
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When conditions for a gas change under constant pressure (and the number of molecules doesn't change), it follows Charles' Law:

\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}  where the temperatures must be measured in Kelvin

To convert from Celsius to Kelvin, add 273, or use the equation:  T_C+273=T_K

For this problem, one must also recall that standard temperature is 0°C (or 273K).

So, T_1 = 273[K], and T_2 = (49.4+273)[K]=322.4[K].

\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}

\dfrac{(1532.7[mL])}{(273[K])}=\dfrac{V_2}{(322.4[K])}

\dfrac{(1532.7[mL])}{(273[K\!\!\!\!\!{-}])}(322.4[K\!\!\!\!\!{-}] )=\dfrac{V_2}{(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})}(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})

1810.04571428[mL]=V_2

Adjusting for significant figures, this gives V_2=1810[mL]

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