Question

If you added 15,000 calories to 2.0 L of water that was at 25.0 degrees C, what temperature would it be at when you finished?

Answer 1

When we finish, the temperature would be 32.5℃

Step-by-step explanation:

Density of water = mass/volume

So,

Mass of water = Density × Volume

`\\\\$=1.0   *  2.0 L$\\\\$=1.0 (g)/(m L) * 2000 m L$\\\\$\quad=2000 g$`

`$Q=m * c * \Delta T$`

where

`\Delta T` = Final T - Initial T

Q is the heat energy in calories

c is the specific heat capacity (for water 1.0 cal/(g℃))

m is the mass of water

plugging in the values

`$15000 \mathrm{Cal}=2000 \mathrm{g} * 1.0 \frac{\mathrm{cal}}{\mathrm{g}^(\circ) \mathrm{C}} * \Delta T$`

`\\$\Delta T=\frac{15000 \mathrm{cal}}{2000 \mathrm{g} * \frac{1.0 \mathrm{cal}}{g^(\circ) \mathrm{C}}}$\\\\$\Delta T=7.5^(\circ) \mathrm{C}$`

Final T = ∆T + Initial T

= 7.5℃ + 25℃ = 32.5℃ (Answer).