Mario uses a hot plate to heat a beaker of 50mL of water. He used a thermometer to measure the temperature of the water. The wat
2 answers:
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Answer:
There are necessaries 35,2g of NH₄NO₃ per 100,0g of water to decrease the temperature of the solution from 25,0°C to 5,0°C
Explanation:
To decrease the temperature of the solution there are necessaries:
4,184J/g°C×(5,0°C-25,0°C)×(100,0g+X) = -Y
8368J + 83,68J/gX = Y <em>(1)</em>
Where x are grams of NH₄NO₃ you need to add and Y is the energy that you need to decrease the heat.
Also, the energy Y will be:
Y = 25700J/mol×X
Y = 321J/g X <em>(2)</em>
Replacing (2) in (1)
8368J + 83,68J/g X = 321J/g X
8363J = 237,32J/gX
<em>X = 35,2g</em>
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Thus, there are necessaries 35,2g of NH₄NO₃ per 100,0g of water to decrease the temperature of the solution from 25,0°C to 5,0°C
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0.208 is the specific heat capacity of the metal.
Explanation:
Given:
mass (m) = 63.5 grams 0R 0.0635 kg
Heat absorbed (q) = 355 Joules
Δ T (change in temperature) = 4.56 degrees or 273.15+4.56 = 268.59 K
cp (specific heat capacity) = ?
the formula used for heat absorbed and to calculate specific heat capacity of a substance will be calculated by using the equation:
q = mc Δ T
c =
c =
= 0.208 J/gm K
specific heat capacity of 0.208 J/gm K
The specific heat capacity is defined as the heat required to raise the temperature of a substance which is 1 gram. The temperature is in Kelvin and energy required is in joules.