Chapter 23, Problem 21A

Calculate the energy absorbed by 20 grams of water that warms from $30°C$ to $90°C$.

To determine

The energy absorbed by 20 grams of water that warms from 30⁰C to 90⁰ C.

Answer to Problem 21A

The heat absorbed is 1200 Calorie.

Explanation of Solution

Given info:

The amount of water m = 20 g

The change in temperature of water

  $\text{ΔT=}\left({\text{90}}^{\text{0}}{\text{C-30}}^{\text{0}}\text{C}\right)={\text{60}}^{\text{0}}\text{C}$

The specific capacity of water C_w= 1 Cal/g ⁰C

Let Q is the heat energy absorbed by water.

Formula used:

  ${\text{Q = mC}}_{\text{w}}\text{ΔT}$

Calculation:

Substituting the given values, we get

  $\begin{array}{l}\text{Q =}\left(\text{20g}\right)\left({\text{1 Cal/g}}^{\text{0}}\text{C}\right)\left({\text{60}}^{\text{0}}\text{C}\right)\\ \text{Q = 1200 Cal}\end{array}$

Conclusion:

Thus, the energy absorbed by 20 g water that warms from 30⁰C to 60⁰C is 20 Calorie.