Question

If 200 grams of water is to be heated from 20.0° C to 100.0° C to make a cup of tea. Howmuch heat was added to the water? (c = 4.18 J/g°C).

Answer 1

66880 J

Explanation:

Given:

Mass (m) = 200 g

Temperature change (ΔT) = 100°C - 20°C = 80°C

Specific heat (c) = 4.18 J/g°C

La energía térmica se calcula mediante la siguiente fórmula:

`\begin{gathered} Q=m\cdot c\cdot\Delta T \\ \\ \text{ We replacing:} \\ \\ Q=200\cdot4.18\cdot80 \\ \\ Q=66880\text{ J} \end{gathered}`

Therefore, a total of 66880 joules was added