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Two hundred grams of a substance requires 0.52 kJ of heat to raise its temperature from 25°C to 45°C. Use the table to identify the substance. q = mC▲T. Mass (m) is in grams. Temperature is in degrees Celsius.


Substance: Specific Heat (C) in joules per gram/degrees centigrade:

Water (ice) 2.05
Iron 0.46
Aluminium 0.90
Gold 0.13
Copper 0.39
Ammonia (liquid) 4.70
Ethanol 2.44
Gasoline 2.22
Water (liquid) 4.18
Water (vapor) 2.08
Air (25 degrees Celsius) 1.01
Oxygen 0.92
Hydrogen 14.30


Question options:

water

gasoline

ammonia

gold

Respuesta :

mixter17
Hi!

The substance is Gold

To know that we clear the equation q=m*C*Δt for the specific heat (C) and solve the equation. Keep in mind that we need to use the heat (0,52 kJ) in Joules since the values from the table are given in Joules. So 0,52 kJ=520 J

[tex] \frac{q}{m*\Delta T}= C= \frac{520 J}{200 g *(45 degC-25degC)}= 0,13 J/gdegC[/tex]

Looking at the table, the substance that has this specific heat is Gold.

Have a nice day!

Answer:a

Explanation:

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