Heat and Energy — ICSE Class 9 Physics Important Questions

Answer: (b) joule.

Heat is a form of energy, so its SI unit is the joule $(\text{J})$(J). The older unit was the calorie, where $1\ \text{cal}=4.2\ \text{J}$1 cal=4.2 J.

Answer: (b) $4^\circ\text{C}$4^.

Due to the anomalous expansion of water, its volume is least (and density greatest) at $4^\circ\text{C}$4^.

Answer: (c) $42000\ \text{J}$42000 J.

$Q=mc\,\Delta T=2\times4200\times5=42000\ \text{J}$Q=mc\,Δ T=2×4200×5=42000 J.

Answer: (b) higher specific heat capacity.

Water's large specific heat capacity makes it heat and cool more slowly than land. This temperature difference sets up convection currents, giving the sea breeze by day and land breeze by night.

Answer: (a) Because of anomalous expansion, water expands as it turns to ice, so the ice occupies more volume and exerts pressure that bursts the pipe. R correctly explains A.

Heat is the total thermal energy possessed by a body due to the motion of its molecules; it is a form of energy measured in joules $(\text{J})$(J) and flows from a hotter to a colder body.

Temperature is the degree of hotness or coldness of a body that decides the direction of heat flow; it is measured in $^\circ\text{C}$^ or kelvin $(\text{K})$(K). Heat is the cause, temperature is the effect.

Water has a very high specific heat capacity $(4200\ \text{J kg}^{-1}\,^\circ\text{C}^{-1})$(4200 J kg^-1\,^^-1).

Car radiator: water can absorb a large quantity of heat from the engine with only a small rise in its own temperature, making it an excellent coolant.

Hot-water bottle: for the same reason, hot water stores a large amount of heat and releases it slowly, so it stays warm for a long time.

Given: $m=0.5\ \text{kg}$m=0.5 kg, initial temperature $=20^\circ\text{C}$=20^, final temperature $=70^\circ\text{C}$=70^, $Q=9000\ \text{J}$Q=9000 J.

$\Delta T=70-20=50^\circ\text{C}$Δ T=70-20=50^

Using $Q=mc\,\Delta T$Q=mc\,Δ T,

$c=\frac{Q}{m\,\Delta T}=\frac{9000}{0.5\times50}=\frac{9000}{25}=360\ \text{J kg}^{-1}\,^\circ\text{C}^{-1}.$c=Q/m\,Δ T=9000/0.5×50=9000/25=360 J kg^-1\,^^-1.

The specific heat capacity of the metal is $360\ \text{J kg}^{-1}\,^\circ\text{C}^{-1}$360 J kg^-1\,^^-1.

Most liquids expand continuously on heating, but water behaves anomalously between $0^\circ\text{C}$0^ and $4^\circ\text{C}$4^. When water is heated from $0^\circ\text{C}$0^ to $4^\circ\text{C}$4^ it contracts instead of expanding; only above $4^\circ\text{C}$4^ does it expand normally. Hence water has maximum density at $4^\circ\text{C}$4^.

Importance: In a freezing pond, the coldest water at the surface turns to ice while the denser water at $4^\circ\text{C}$4^ sinks to the bottom. Ice, being a poor conductor, floats and insulates the water below, which stays at about $4^\circ\text{C}$4^. Fish and other aquatic animals survive in this liquid water beneath the ice.

By the principle of calorimetry, heat lost by iron $=$= heat gained by water.

Heat gained by water:
$Q_w=m_w c_w \Delta T_w=0.5\times4200\times(22-20)=0.5\times4200\times2=4200\ \text{J}.$Q_w=m_w c_w Δ T_w=0.5×4200×(22-20)=0.5×4200×2=4200 J.

Heat lost by iron:
$Q_i=m_i c_i \Delta T_i=0.2\times c_i\times(100-22)=0.2\times c_i\times78=15.6\,c_i.$Q_i=m_i c_i Δ T_i=0.2× c_i×(100-22)=0.2× c_i×78=15.6\,c_i.

Equating $Q_i=Q_w$Q_i=Q_w:
$15.6\,c_i=4200\ \Rightarrow\ c_i=\frac{4200}{15.6}\approx269.2\ \text{J kg}^{-1}\,^\circ\text{C}^{-1}.$15.6\,c_i=4200 c_i=4200/15.6269.2 J kg^-1\,^^-1.

The specific heat capacity of iron is about $269\ \text{J kg}^{-1}\,^\circ\text{C}^{-1}$269 J kg^-1\,^^-1.

(a) Specific heat capacity is the quantity of heat required to raise the temperature of unit mass $(1\ \text{kg})$(1 kg) of a substance by $1^\circ\text{C}$1^ (or $1\ \text{K}$1 K). SI unit: $\text{J kg}^{-1}\,^\circ\text{C}^{-1}$J kg^-1\,^^-1 (or $\text{J kg}^{-1}\,\text{K}^{-1}$J kg^-1\,K^-1).

(b) Renewable: solar energy, wind energy (also hydro, tidal, geothermal, biomass). Non-renewable: coal, petroleum (also natural gas, nuclear fuels).

(c) $\Delta T=80-30=50^\circ\text{C}$Δ T=80-30=50^.
$Q=mc\,\Delta T=3\times4200\times50=630000\ \text{J}=6.3\times10^{5}\ \text{J}.$Q=mc\,Δ T=3×4200×50=630000 J=6.3×10^5 J.

The water releases $6.3\times10^{5}\ \text{J}$6.3×10^5 J of heat energy.

(a) The greenhouse effect is the trapping of the Sun's infrared (heat) radiation near the Earth's surface by certain atmospheric gases, which keeps the Earth warm. Greenhouse gases: carbon dioxide $(CO_2)$(CO_2) and methane $(CH_4)$(CH_4) (also water vapour, nitrous oxide, CFCs).

(b) Harmful effects: (i) melting of polar ice caps and glaciers causing a rise in sea level and flooding of coastal areas; (ii) disturbance of climate and rainfall patterns, harming crops and wildlife.

(c) Heat needed:
$Q=mc\,\Delta T=1\times4200\times(75-25)=4200\times50=210000\ \text{J}.$Q=mc\,Δ T=1×4200×(75-25)=4200×50=210000 J.
Time taken:
$t=\frac{Q}{\text{power}}=\frac{210000}{2100}=100\ \text{s}.$t=Q/power=210000/2100=100 s.

It takes $100\ \text{s}$100 s (about $1\ \text{min}\ 40\ \text{s}$1 min 40 s).

(i) The principle of calorimetry (principle of mixtures): when two bodies at different temperatures are mixed, heat lost by the hot body equals heat gained by the cold body, provided no heat escapes.

(ii) Heat lost by X $=$= heat gained by Y:
$0.4\times4200\times(60-\theta)=0.6\times4200\times(\theta-20).$0.4×4200×(60-)=0.6×4200×(-20).

(iii) Cancel $4200$4200:
$0.4(60-\theta)=0.6(\theta-20)$0.4(60-)=0.6(-20)
$24-0.4\theta=0.6\theta-12$24-0.4=0.6-12
$24+12=0.6\theta+0.4\theta$24+12=0.6+0.4
$36=\theta\ \Rightarrow\ \theta=36^\circ\text{C}.$36= =36^.

(iv) So that no heat is lost to or gained from the surroundings; otherwise heat lost by the hot sample would not exactly equal heat gained by the cold sample, and the calculated temperature would be wrong.