Chapter 12.2, Problem 20PP

To determine

The amount of thermal energy absorbed.

Answer to Problem 20PP

The amount of thermal energy absorbed is $6.69\times {10}^4\text{ J}$.

Explanation of Solution

Given:

The temperature of $2.00\times {10}^2 \text{g}$ water is changed from $60°\text{C}$ to $140°\text{C}$ by heating.

Formula Used:

When temperature of an object is changed from

  $T_A$ (in Kelvin) to $T_B$ , there is always a gain or loss of thermal energy. This energy will be dependent on the temperature change, mass of the object $m$ and specific heat $C$ , and is given by the expression,

  $Q=mC\left(T_B-T_A\right)$ ..... (1)

Calculation:

Here, the given substance is water which has a mass of $m\text{=2}.00\times \text{1}{0}^{\text{2}}-\text{g}$ . The temperature of water is changed from 60⁰C to 140⁰C. The specific heat of water is $C=4180\text{ J/kg}\text{.K}$ .

The initial temperature of water in Kelvin can be calculated as,

  $\begin{array}{c}{T}_A=60+273\\ =333\text{ K}\end{array}$

Similarly, the final temperature of iron in Kelvin is,

  $\begin{array}{c}{T}_A=140+273\\ =413\text{ K}\end{array}$

Substituting these parameters in (1), the thermal energy absorbed to achieve this temperature change is,

  $\begin{array}{c}Q=0.2\times 4180\left(413-333\right)\\ =6.69\times {10}^4\text{ J}\end{array}$

Conclusion:

Thus, the amount of thermal energy absorbed to raise the temperature of water for the given range is $6.69\times {10}^4\text{ J}$ .