Question

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

Answer 1

23.95 °C

Step-by-step 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