Question

How many joules of energy are required to increase the temperature of 750 grams of water from 20.5 degrees Celsius to 35.8 degrees Celsius?

Answer 1

To calculate the energy required to increase the temperature of water, we can use the formula for heat transfer:

\[
Q = mc\Delta T
\]

where:
- \( Q \) is the amount of heat energy (in joules),
- \( m \) is the mass of the substance (in kilograms),
- \( c \) is the specific heat capacity of the substance (for water, it is approximately \( 4.18 \, \text{J/g}^\circ\text{C} \)),
- \( \Delta T \) is the change in temperature (in degrees Celsius).

Given:
- Mass \( m = 750 \, \text{g} \)
- Initial temperature \( T_i = 20.5 \,^\circ\text{C} \)
- Final temperature \( T_f = 35.8 \,^\circ\text{C} \)

First, we calculate the change in temperature (\( \Delta T \)):

\[
\Delta T = T_f - T_i = 35.8 \,^\circ\text{C} - 20.5 \,^\circ\text{C} = 15.3 \,^\circ\text{C}
\]

Now, we can substitute the values into the formula. Note that we need to convert the mass to kilograms for consistency with the specific heat capacity:

\[
m = 750 \, \text{g} = 0.750 \, \text{kg}
\]

Now we can calculate \( Q \):

\[
Q = 750 \, \text{g} \times 4.18 \, \text{J/g}^\circ\text{C} \times 15.3 \,^\circ\text{C}
\]

Calculating it step by step:

\[
Q = 750 \times 4.18 \times 15.3
\]
\[
= 750 \times 64.074 = 48055.5 \, \text{J}
\]

Therefore, the energy required to increase the temperature of 750 grams of water from 20.5 degrees Celsius to 35.8 degrees Celsius is approximately:

\[
\boxed{48055.5} \, \text{J}
\]