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if .709 j of heat is added to water and cause the temperature to go up by .036 degrees C what mass of water is present

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Answer:

0.00471 grams H₂O

Step-by-step explanation:

To determine the mass, you need to use the following equation:

Q = mcΔT

In this equation,

-----> Q = energy/heat (J)

-----> m = mass (g)

-----> c = specific heat capacity (J/g°C)

-----> ΔT = temperature change (°C)

The specific heat capacity of water is 4182 J/g°C. You can plug the given values into the equation and simplify to isolate "c".

Q = 0.709 J c = 4182 J/g°C

m = ? g ΔT = 0.036 °C

Q = mcΔT <----- Equation

0.709 J = m(4182 J/g°C)(0.036 °C) <----- Insert values

0.709 J = m(150.552) <----- Multiply 4182 and 0.036

0.00471 = m <----- Divide both sides by 150.552

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