If 0.507J of heat leads to a 0.007 degree C change in water, what mass is present?


0.0337 g

17.3 g

0.239 g

0.300 g

Answers

Answer 1

Answer:

17.3 g

Explanation:

Given the following data;

Quantity of heat, Q = 0.507 JTemperature = 0.007°CSpecific heat capacity of water = 4.2 J/g°C

Mathematically, Heat capacity is given by the formula;

[tex] Q = MCT [/tex]

Where;

Q represents the heat capacity or quantity of heat. M represents the mass of an object. C represents the specific heat capacity of water. T represents the temperature.

Making "M" the subject of formula, we have;

[tex] Mass, \ m = \frac {Q}{CT} [/tex]

Substituting the values into the formula, we have;

[tex] Mass, \ m = \frac {0.507}{0.007*4.2} [/tex]

[tex] Mass, \ m = \frac {0.507}{0.0294} [/tex]

Mass, m = 17.3 grams


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Answers

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