Question

18. A 35.0 G piece of metal wire is heated and the temperature changes from 21°C to 52°C the specific heat of the metal is 0.900 J/G degrees Celsius how much energy was required to heat the metal please show work

Answer 1

Answer:

$\boxed {\boxed {\sf 976.5 \ Joules}}$

Explanation:

We are asked to find the heat energy. Since we are given the mass, specific heat, and change in temperature, we should use this formula for heat:

$q=mc\Delta T$

Where m is the mass, c is the specific heat, and ΔT is the change in temperature.

We know this is a 35.0 gram piece of wire, the temperature changes from 21 °C to 52°C and the specific heat is 0.900 J/ g °C.

Therefore,

• m= 35.0 g
• c= 0.900 J/g °C
• ΔT= 52 °C - 21°C= 31°C

Substitute these values into the formula.

$q= (35.0 \ g)(0.900 \ J/g \ \textdegree C)(31 \textdegree C)$

Multiply the first two values. The grams will cancel each other out.

$q=31.5 \ J/ \textdegree C (31 \textdegree C)$

Multiply again. This time, the degrees Celsius cancel each other out, so the final units are Joules.

$q=976.5 \ J$

976.5 Joules of energy were required to heat the piece of metal.