Chapter U5.97, Problem 4E

Interpretation Introduction

Interpretation:

The method that warm an inflatable pool of a child more needs to be identified.

Concept introduction:

The energy contained in the motion and vibration of its molecules is known as the thermal energy of an object.

A physical property of matter which quantitatively shows hot and cold is known as temperature.

Explanation of Solution

First, find out the amount of heat required in both cases:

So, applying the formula:

$q=mC\Delta T$

Here,

q = heat

m = mass

C = specific heat

$\text{Δ}T$ = change in temperature.

Case 1:

m = 500 g

C = 1(cal/g°C)

$\text{Δ}T$ = (50 − 20)°C

$q=mC\Delta T$

Putting the values,

$\begin{array}{l}$ q=500\times 1\frac{\text{cal}}{{\text{g}}^{\text{0}}\text{C}}\times {(50-20)}^0\text{C}$q=50\text{0 g}\times 1\frac{\text{cal}}{{\text{g}}^{\text{0}}\text{C}}\times {30}^0\text{C}\\ q=15000\text{ cal}\end{array}$

Similarly, for case 2:

m = 100 g

C = 1(cal/g°C)

$\text{Δ}T$ = (95 − 20)°C

$q=mC\Delta T$

Putting the values,

$\begin{array}{l}$ q=100\times 1\frac{\text{cal}}{{\text{g}}^{\text{0}}\text{C}}\times {(95-20)}^{\text{0}}\text{C}$q=10\text{0 g}\times 1\frac{\text{cal}}{{\text{g}}^{\text{0}}\text{C}}\times {75}^0\text{C}\\ q=7500\text{ cal}\end{array}$

Thus, upon adding 500 g of water at 50°C, the pool will heat more because more amount of heat is transferred.