Answer:
[tex]\boxed {\boxed {\sf C. \ 1935 \ J}}[/tex]
Explanation:
The equation for this problem is:
[tex]q=mc\Delta T[/tex]
where m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
The mass is 3 kilograms, but the specific heat capacity includes grams in the units. Convert kilograms to grams. There are 1000 grams in 1 kilogram.
- [tex]\frac {1000 \ g}{1 \ kg}[/tex]
- [tex]3 \ kg *\frac {1000 \ g}{1 \ kg}[/tex]
- [tex]3 *1000 \ g = 3000 \ g[/tex]
The specific heat capacity for lead is found on the table. It is 0.129 J/g°C.
Let's find the change in temperature. It is raised from 15 °C to 20 °C.
- [tex]\Delta T= final \ temperature - initial \ temperature \\\Delta T= 20 \textdegree C - 15 \textdegree C\\\Delta T= 5 \textdegree C[/tex]
Now we know every value.
- m= 3000 g
- c= 0.129 J/g°C
- ΔT= 5 °C
Substitute the values into the formula.
[tex]q= (3000 \ g)( 0.129 \ J/g \textdegree C)(5 \textdegree C)[/tex]
Multiply the first 2 numbers together. The units of grams cancel.
[tex]q= (387 \ J/ \textdegree C )(5 \textdegree C)[/tex]
Multiply again. This time the units of degrees Celsius cancel.
[tex]q= 1935 \ J[/tex]
1935 Joules of energy are required and choice C is correct.
