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

23.95 °C

We are given;

We are required to calculate the 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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