Question

An sample of water with a mass of 123.00 kg is heated from 25 C to 97 C. If the specific heat of water is 1 J/kg K, then how much energy is required?A. 11.93 kJB. 15.01 kJC. 123.00 kJD. 8.86 kJ

Answer 1

Answer

D. 8.86 kJ

Explanation

Given that:

Mass of water, m = 123.00 kg

Initial temperature, T₁ = 25 °C = (25 °C + 273) = 298 K

Final temperature, T₂ = 97 °C = (97 °C + 273) = 370 K

Specific heat of water, c = 1 J/kg K

What to find:

The amount of energy (Q) required to cause the temperature rise of the mass of water.

Step-by-step solution:

The amount of energy (Q) required to cause the temperature rise of the mass of water can be calculated using the formula below:

$Q=mc\Delta T$

Note that ΔT = T₂ - T₁ = 370 K - 298 K = 72 K

Therefore, the energy require will be:

$\begin{gathered} Q=123\text{ }kg\times1\text{ }J\text{/}kgK\times72\text{ }K \\ \\ Q=8,856\text{ }J \\ \\ Q\approx8.86\text{ }kJ \end{gathered}$