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A water heater can generate 32000 kJ/h. How much water can it heat from \(12{\rm{\circ C}}\)to \(14{\rm{\circ C}}\)

Short Answer

Expert verified

It can heat \(254.82\,\,{\rm{kg}}\) water per hour.

Step by step solution

01

Definition of heat capacity

The heat capacity of a system represents the heat energy required to raise the temperature of the system.The heat capacity of the system is dependent on the mass and temperature of the system. In other words, the mass and temperature of the system are directly proportional to the heat capacity of the system.

The expression for the heat generated by the water heater per hour is

\(Q = mc\left( {{T_{\rm{f}}} - {T_{\rm{i}}}} \right)\).

Here, \(m\)is mass of water, \(c\)is the specific heat capacity of water, \({T_{\rm{f}}}\)is the final temperature of water, and \({T_{\rm{i}}}\) is the initial temperature of water.

02

Given information

The final temperature of the water is\({T_{\rm{f}}} = 42{\rm{\circ C}}\).

The initial temperature of the water is\({T_{\rm{i}}} = 12{\rm{\circ C}}\).

The heat capacity of water is \(Q = 32000\,{\rm{kJ}}\).

03

Calculation of the mass of the water per hour

The mass of water can be calculated as shown below:

\(\begin{array}{c}m = \frac{Q}{{c\left( {{T_{\rm{f}}} - {T_{\rm{i}}}} \right)}}\\ = \frac{{\left( {32000\,{\rm{kJ}}} \right)\left( {\frac{{\left( {{{10}3}\,{\rm{J}}} \right)}}{{1\,{\rm{kJ}}}}} \right)}}{{\left( {4816\,\,{{\rm{J}} \mathord{\left/{\vphantom {{\rm{J}} {{\rm{kg}}}}} \right.\\} {{\rm{kg}}}} \cdot {\rm{\circ C}}} \right)\left( {\left( {42\,\circ {\rm{C}}} \right) - \left( {12\,\circ {\rm{C}}} \right)} \right)}}\\ = 254.82\,{\rm{kg}}\end{array}\)

Hence, the mass of water heated per hour is \(254.82\,\,{\rm{kg}}\).

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Most popular questions from this chapter

(II)The 1.20-kg head of a hammer has a speed of 7.5 m/s just before it strikes a nail (Fig. 14鈥17) and is brought to rest. Estimate the temperature rise of a 14-g iron nail generated by eight such hammer blows done in quick succession. Assume the nail absorbs all the energy.

A leaf of area \({\bf{40}}\;{\bf{c}}{{\bf{m}}^{\bf{2}}}\) and mass \({\bf{4}}{\bf{.5 \times 1}}{{\bf{0}}^{{\bf{ - 4}}}}\;{\bf{kg}}\) directly faces the Sun on a clear day. The leaf has an emissivity of 0.85 and a specific heat of \({\bf{0}}{\bf{.80}}\;{\bf{kcal/kg}} \cdot {\bf{K}}\) (a) Estimate the energy absorbed per second by the leaf from the Sun, and then (b) estimate the rate of rise of the leaf鈥檚 temperature. (c) Will the temperature rise continue for hours? Why or why not? (d) Calculate the temperature the leaf would reach if it lost all its heat by radiation to the surroundings at 24掳C. (e) In what other ways can the heat be dissipated by the leaf?

(a) Estimate the total power radiated into space by the Sun, assuming it to be a perfect emitter at \(T = 5500\;{\rm{K}}\). The Sun鈥檚 radius is \({\bf{7 \times 1}}{{\bf{0}}{\bf{8}}}\;{\bf{m}}\). (b) From this, determine the power per unit area arriving at the Earth, away \({\bf{1}}{\bf{.5 \times 1}}{{\bf{0}}{{\bf{11}}}}\;{\bf{m}}\) (Fig. 14鈥20).

FIGURE 14-20

Problem 47.

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