(a) Calculate the energy available when a 106 kg parcel of water is cooled by $4 ^ { \circ } \mathrm { C }$.
(b) If this energy is used (at an efficiency of $100 \%$ ) to lift the parcel of water against gravity, what would be its elevation?
(c) How long would this energy satisfy the average per-capita primary energy needs in Norway?

Question

Quizplus · Accepted Answer

The Answer is (a) The thermal energy is

E = m C \Delta T

Using the values

\begin{array} { l } m = 10 ^ { 6 } \mathrm {~kg} \C = 4180 \mathrm {~J} / \left( \mathrm { kg } \cdot { } ^ { \circ } \mathrm { C } ight) \\Delta T = 4 ^ { \circ } \mathrm { C }\end{array}

gives

E = \left( 10 ^ { 6 } \mathrm {~kg} ight) imes \left( 4180 \mathrm {~J} / \left( \mathrm { kg } \cdot { } ^ { \circ } \mathrm { C } ight) ight) imes \left( 4 ^ { \circ } \mathrm { C } ight) = 1.672 imes 10 ^ { 10 } \mathrm {~J}

(b) The gravitational potential energy is

E = m g h

Solving for h gives

\left. h = E / ( m g ) = \left( 1.672 imes 10 ^ { 10 } \mathrm {~J} ight) / \left( 10 ^ { 6 } \mathrm {~kg} ight) imes \left( 9.8 \mathrm {~m} / \mathrm { s } ^ { 2 } ight) ight) = 1.71 imes 10 ^ { 3 } \mathrm {~m}

(c) From Figure

2.4

for Norway the per capita energy use is about

440 \mathrm { GJ } / \mathrm { y } = 4.4 imes 10 ^ { 11 } \mathrm {~J} / \mathrm { y }

. From part (a) the energy would last for

\left( 1.672 imes 10 ^ { 10 } \mathrm {~J} ight) / \left( 4.4 imes 10 ^ { 11 } \mathrm {~J} / \mathrm { y } ight) = 0.038 \mathrm { y }

or about 2 weeks.

(A) Calculate the Energy Available When a 106 Kg Parcel 4∘C4 ^ { \circ } \mathrm { C }4∘C

(A) Calculate the Energy Available When a 106 Kg Parcel $4 ^ { \circ } \mathrm { C }$