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What would the expected temperature change be (in Fahrenheit) ida 0.5 gran sample of water released 50.1 J of heat energy? The specific heat of liquid water is 4.184 J/g-C

Respuesta :

nkirotedoreen46

Answer:

23.95 °C

Explanation:

We are given;

  • Mass of the sample is 0.5 gram
  • Quantity of heat released as 50.1 Joules
  • Specific heat capacity is 4.184 J/g°C

We are required to calculate the change in temperature;

  • Quantity of heat absorbed is given by the formula;
  • Q = mass × specific heat capacity × Change in temperature

That is, Q = mcΔT

Rearranging the formula;

ΔT = Q ÷ mc

Therefore;

ΔT = 50.1 J ÷ (0.5 g × 4.184 J/g°C)

    = 23.95 °C

Therefore, the expected change in temperature is 23.95 °C

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