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

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