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notsponge [240]
3 years ago
6

The specific heat of aluminum is 0.125 cal/g °C. If 12.5 grams of aluminum were heated from 20.0 °C to 100.0 °C, calories of hea

t energy would be absorbed by the aluminum.
Chemistry
2 answers:
WINSTONCH [101]3 years ago
7 0

Answer:

Q=125cal

Explanation:

Hello,

In this case, this could be determined via the following equation for the determination of the absorbed heat by the aluminium, considering the heat capacity, the change in the temperature and the heated mass:

Q=mCp\Delta T\\Q=mCp(T_2-T_1)\\Q=12.5g*0.125\frac{cal}{g*^oC}*(100.0-20.0)^oC\\ Q=125cal

Best regards.

klemol [59]3 years ago
4 0
Aluminum does not undergo any phase change from 20.0°C to 100<span>.0°C, therefore we only consider sensible heat change, which is calculate as
</span>q = mc(dT)q = (12.5 g)(0.125 cal/g-°C)(100 - 20)<span>°C = 125 cal
</span>
Therefore, 125 calories of heat are absorbed by the aluminum.
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Define mass number
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protons plus neutrons

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Which (3 of the 4) quantities must be conserved in all chemical reactions?
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3 years ago
Para formar bronce, se mezclan 150g de cobre a 1100°C y 35g de estaño a 560°C. Determine la temperatura final del sistema.
jek_recluse [69]

Answer:

La temperatura final del sistema es 1029,346 °C.

Explanation:

Asumamos que el sistema conformado por el cobre y el estaño no tiene interacciones con sus alrededores. Por la Primera Ley de la Termodinámica, el cobre cede calor al estaño con tal de alcanzar el equilibrio térmico. El cobre se encuentra inicialmente en su punto de fusión, mientras que el estaño está por encima de ese punto, de modo que la transferencia de calor es esencialmente sensible:

m_{Cu}\cdot c_{Cu}\cdot (T-T_{Cu}) = m_{Sn}\cdot c_{Sn}\cdot (T_{Sn}-T)

(m_{Cu}\cdot c_{Cu} + m_{Sn}\cdot c_{Sn})\cdot T = m_{Sn}\cdot c_{Sn}\cdot T_{Sn} + m_{Cu}\cdot c_{Cu}\cdot T_{Cu}

T = \frac{m_{Sn}\cdot c_{Sn}\cdot T_{Sn}+m_{Cu}\cdot c_{Cu}\cdot T_{Cu}}{m_{Cu}\cdot c_{Cu}+m_{Sn}\cdot c_{Sn}} (1)

Donde:

m_{Sn} - Masa del estaño, en gramos.

m_{Cu} - Masa del cobre, en gramos.

c_{Sn} - Calor específico del estaño, en calorías por gramo-grados Celsius.

c_{Cu} - Calor específico del cobre, en calorías por gramo-grados Celsius.

T_{Sn} - Temperatura inicial del estaño, en grados Celsius.

T_{Cu} - Temperatura inicial del cobre, en grados Celsius.

Si sabemos que m_{Cu} = 150\,g, m_{Sn} = 35\,g, c_{Cu} = 0,093\,\frac{cal}{g\cdot ^{\circ}C}, c_{Sn} = 0,060\,\frac{cal}{g\cdot ^{\circ}C}, T_{Sn} = 560\,^{\circ}C y T_{Cu} = 1100\,^{\circ}C, entonces la temperatura final del sistema es:

T = \frac{(35\,g)\cdot \left(0,060\,\frac{cal}{g\cdot ^{\circ}C} \right)\cdot (560\,^{\circ}C)+(150\,g)\cdot \left(0,093\,\frac{cal}{g\cdot ^{\circ}C} \right)\cdot (1100\,^{\circ}C)}{(35\,g)\cdot \left(0,060\,\frac{cal}{g\cdot ^{\circ}C} \right)+(150\,g)\cdot \left(0,093\,\frac{cal}{g\cdot ^{\circ}C} \right)}

T = 1029,346\,^{\circ}C

La temperatura final del sistema es 1029,346 °C.

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