Question

How much energy is required to raise the temperature of 3 kg of lead from 15°C to 20°C? Use the table below and this equation: Q = MCAT.
The question is written right above the table given.

Answer 1

Answer:

1935J

Explanation:

Answer 2

Answer:

$\boxed {\boxed {\sf C. \ 1935 \ J}}$

Explanation:

The equation for this problem is:

$q=mc\Delta T$

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.

• $\frac {1000 \ g}{1 \ kg}$

• $3 \ kg *\frac {1000 \ g}{1 \ kg}$

• $3 *1000 \ g = 3000 \ g$

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.

• $\Delta T= final \ temperature - initial \ temperature \\\Delta T= 20 \textdegree C - 15 \textdegree C\\\Delta T= 5 \textdegree C$

Now we know every value.

• m= 3000 g
• c= 0.129 J/g°C
• ΔT= 5 °C

Substitute the values into the formula.

$q= (3000 \ g)( 0.129 \ J/g \textdegree C)(5 \textdegree C)$

Multiply the first 2 numbers together. The units of grams cancel.

$q= (387 \ J/ \textdegree C )(5 \textdegree C)$

Multiply again. This time the units of degrees Celsius cancel.

$q= 1935 \ J$

1935 Joules of energy are required and choice C is correct.