Q. 3.99 - An Introduction to General, Organic, and Biological Chemistry

Question

In a large building, oil is used in a steam boiler heating system. The combustion of $1.0lb$ of oil provides $2.4 \times {10}^{7}J$. $(3.4,3.6)$

$\left(a\right)$How many kilograms of oil are needed to heat $150 kg$ of water from ${22}^{\circ }C$ to ${100}^{\circ }C$ ?

$\left(b\right)$How many kilograms of oil are needed to change $150 kg$ of water to steam at ${100}^{\circ }C$ ?

Step-by-Step Solution

Answer

Part $\left(a\right)$

$\left(a\right)$$0.93 kg$ of oil is required to heat $150kg$of water from ${22}^{\circ }C$ to ${100}^{\circ }C$.

Part $\left(b\right)$

$\left(b\right)$$6.43 kg$of oil is required to heat  $150 kg$ of steam water at${100}^{\circ }C.$

Step: Given information: (Part a)

Given are:$1.0lb$ of oil and $2.4\times {10}^{7}J$ and $150kg$of water and temperature is${22}^{\circ }C-{100}^{\circ }C.$

To find the oil in kilogram.

Step: 2 Finding heat: (Part a)

The mass of water is $=150\mathrm{kg}$

$=150\times 1000\mathrm{g}$

The change in temperature as ${100}^{\circ }\mathrm{C}-{22}^{\circ }\mathrm{C}={78}^{\circ }\mathrm{C}$

$\text{ Heat required }=\text{ mass }\times \text{ temperature change }\times \text{ specific heat }$

$Heat=150\times 1000\mathrm{g}\times {78}^{\circ }\mathrm{C}\times 4.184\mathrm{J}/{\mathrm{g}}^{\circ }\mathrm{C}$

As aresult,oil in kilogram is $Heat=4.895\times {10}^7\mathrm{J}.$

Step: 3 Finding oil in kilogram: (Part a)

Calculating pounds of oil as 

$\text{ pounds of oil required }=\frac{\text{ Heat required }}{\text{ Heat obtained from }1\mathrm{lb}}$

$\text{ pounds of oil required }=\frac{4.895\times {10}^7\mathrm{J}}{2.4\times {10}^7\mathrm{J}}$

$\text{ pounds of oil required }=2.04lb.$

Where,$1kg=2.04lb$

so,$2.04\mathrm{lb}=\frac{2.04}{2.20}\mathrm{kg}$

As a result,oil need in kilogram as $0.93kg.$

Step: 4 Given information: (Part b)

Given are: $150g$ of water and steam is ${100}^{\circ }C.$

To find the oil in kilogram.

Step: 5 Finding oil in kilogram: (Part b)

Mass of water is  $150kg$

Heat vapourization of water is  $2260 kJ/kg$

$\text{Heat }=\text{ Mass }\times \text{ heat of vaporization }$

$Heat=150\mathrm{kg}\times 2260\mathrm{kJ}/\mathrm{kg}$

$Heat==3.4\times {10}^5\mathrm{kJ}$

$Pounds of oil=\frac{3.4\times {10}^5\mathrm{kJ}}{2.4\times {10}^4\mathrm{kJ}/\mathrm{lb}}$

$Pounds of oil =1.413\times {10}^1\mathrm{lb}$

$1.413\times {10}^{\mathrm{l}}\mathrm{lb}=\frac{1.413\times {10}^{\mathrm{l}}}{2.20}\mathrm{kg}$

As a result,the oil in kilogram as $6.4\mathrm{kg}.$