Question

What amount of heat is released when the temperature of 450.0 g of a

substance drops by 7.1 °C? Assume that the specific heat = 1.264 J/g.°C

Answer 1

`\boxed {\boxed {\sf 4038.48 \ Joules}}`

Step-by-step explanation:

Since we are given the mass, specific heat, and change in temperature, we should use this formula for heat:

`q=mc\Delta T`

The substance's mass is 450.0 grams, the specific heat is 1.264 J/g°C, and the change in temperature is 7.1 °C.

`m= 450.0 \ g \\c= 1.264 \ J/g \textdegree C\\\Delta T= 7.1 \ \textdegree C`

Substitute the values into the formula.

`q= (450.0 \ g)(1.264 \ J/g \textdegree C)(7.1 \ \textdegree C)`

Multiply the first 2 values together. The grams will cancel out.

`q= 568.8 \ J/ \textdegree C (7.1 \ \textdegree C)`

Multiply again. This time, the degrees Celsius cancel out.

`q= 4038.48 \ J`

4038.48 Joules of heat energy are released.