Question

If you added 45,000 calories to water that was at 25 degrees C, and the ending temperature was 35 degrees C, how much water did you have (in L)?

Answer 1

4.5 L water we have in litres (L).

Step-by-step explanation:

`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

`\\$45000 \mathrm{cal}=m * 1.0 \frac{\mathrm{cal}}{\mathrm{g}^(\circ) \mathrm{C}} *\left(35^(\circ) \mathrm{C}-25^(\circ) \mathrm{C}\right)$\\\\$45000 \mathrm{cal}=m * 1.0 \frac{\mathrm{cal}}{\mathrm{g}^(\circ) \mathrm{C}} * 10^(\circ) \mathrm{C}$\\\\$m=\frac{45000 \mathrm{cal}}{1.0 \frac{\mathrm{cal}}{\mathrm{g}^(\circ) \mathrm{C}} * 10^(\circ) \mathrm{C}}$\\\\$m=4500 \mathrm{g}$\\\\Density of water $=\frac{\text { mass }}{\text { volume }}$`

So,

Volume of water = mass/density

`\\\\=\frac{4500 \mathrm{g}}{\frac{1.09}{\mathrm{mL}}}=4500 \mathrm{mL}$$`

=4.5 L (Answer)