Question

If 350 grams of lemon is to be heated from 40.0° C to 110.0° C to make a cup of tea. Howmuch heat was added to the lemon? (c = 4.44 J/g°C).

Answer 1

108780 J

Explanation:

Given:

Mass (m) = 350 g

Temperature change (ΔT)= 110°C - 40°C = 70°C

Specific heat (c) = 4.44 J/g°C

Using the following formula we calculate the added heat:

`\begin{gathered} Q=m\cdot c\cdot\Delta T \\ \\ \text{ We replacing} \\ \\ Q=350\cdot4.44\cdot70 \\ \\ Q=108780\text{ J} \end{gathered}`

The amount of heat added is 108780 joules.