Calorimetry — ICSE Class 10 Physics Important Questions

Answer: (b) $\text{J kg}^{-1}\,^\circ\text{C}^{-1}$J kg^-1\,^^-1.

Specific heat capacity $c=\dfrac{Q}{m\,\Delta T}$c=Q/m\,Δ T, so its unit is $\dfrac{\text{J}}{\text{kg}\cdot{}^\circ\text{C}}=\text{J kg}^{-1}\,^\circ\text{C}^{-1}$Jkg·^=J kg^-1\,^^-1. The unit $\text{J }^\circ\text{C}^{-1}$J ^^-1 is that of heat capacity.

Answer: (b) $3360\,\text{J}$3360\,J.

$Q=mL=10\text{ g}\times336\text{ J g}^{-1}=3360\text{ J}$Q=mL=10 g×336 J g^-1=3360 J. The temperature stays at $0\,^\circ\text{C}$0\,^ during melting, so only latent heat is involved.

Answer: (b) high specific heat capacity.

Water's high specific heat capacity ($4200\text{ J kg}^{-1}\,^\circ\text{C}^{-1}$4200 J kg^-1\,^^-1) lets it absorb a large amount of heat for a small rise in temperature, making it an excellent coolant.

Answer: (b) $1:2$1:2.

For equal $m$m and $Q$Q, $\Delta T=\dfrac{Q}{mc}\propto\dfrac{1}{c}$Δ T=Q/mc1/c. So $\dfrac{\Delta T_{water}}{\Delta T_{liquid}}=\dfrac{c_{liquid}}{c_{water}}=\dfrac{2100}{4200}=\dfrac{1}{2}$Δ T_waterΔ T_liquid=c_liquidc_water=2100/4200=1/2.

Answer: (a) While melting, the temperature stays constant, and this is because the supplied heat (latent heat) is used to break the intermolecular bonds and change the state, not to increase kinetic energy/temperature. R correctly explains A.

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

$c=\dfrac{Q}{m\,\Delta T}.$c=Q/m\,Δ T.

Its SI unit is $\text{J kg}^{-1}\,^\circ\text{C}^{-1}$J kg^-1\,^^-1 (or $\text{J kg}^{-1}\text{ K}^{-1}$J kg^-1 K^-1).

Water has a very high specific heat capacity. Therefore it warms up and cools down much more slowly than land.

During the day and in summer, the sea absorbs a large amount of heat with only a small rise in temperature; at night and in winter it releases this heat slowly. This prevents extremes of temperature, so coastal towns have a moderate (equable) climate with neither very hot days nor very cold nights.

Given: $m=2\text{ kg}$m=2 kg.

Step 1 — heating water $30\,^\circ\text{C}\to100\,^\circ\text{C}$30\,^100\,^:
$Q_1=mc\,\Delta T=2\times4200\times(100-30)=2\times4200\times70=588000\text{ J}.$Q_1=mc\,Δ T=2×4200×(100-30)=2×4200×70=588000 J.

Step 2 — converting water at $100\,^\circ\text{C}$100\,^ to steam:
$Q_2=mL=2\times2260000=4520000\text{ J}.$Q_2=mL=2×2260000=4520000 J.

Total heat:
$Q=Q_1+Q_2=588000+4520000=5108000\text{ J}=5.108\times10^{6}\text{ J}.$Q=Q_1+Q_2=588000+4520000=5108000 J=5.108×10^6 J.

Let the final temperature be $\theta$. By the principle of calorimetry, heat lost by hot water = heat gained by cold water.

Heat lost by hot water: $Q_{lost}=0.5\times c\times(80-\theta)$Q_lost=0.5× c×(80-).

Heat gained by cold water: $Q_{gained}=0.3\times c\times(\theta-20)$Q_gained=0.3× c×(-20).

Equating (the $c$c cancels):
$0.5(80-\theta)=0.3(\theta-20)$0.5(80-)=0.3(-20)
$40-0.5\theta=0.3\theta-6$40-0.5=0.3-6
$46=0.8\theta$46=0.8
$\theta=57.5\,^\circ\text{C}.$=57.5\,^.

The final temperature of the mixture is $57.5\,^\circ\text{C}$57.5\,^.

When hot water is placed in a calorimeter, the calorimeter (and stirrer) also gain heat as their temperature rises. If this heat is ignored, the calculated heat exchange (and hence any specific heat found from it) will be in error, because part of the heat lost by the hot body actually goes into warming the calorimeter, not only the other liquid.

Hence the correct heat balance is:
$\text{heat lost by hot body}=\text{heat gained by liquid}+\text{heat gained by calorimeter}.$heat lost by hot body=heat gained by liquid+heat gained by calorimeter.

Water equivalent of a calorimeter is the mass of water that would absorb the same amount of heat as the calorimeter for the same rise in temperature. If the calorimeter has mass $m$m and specific heat capacity $c$c, its water equivalent is
$W=\dfrac{mc}{c_{water}}.$W=mcc_water.

Let the final temperature be $\theta\,^\circ\text{C}$\,^ (assume all ice melts and mixture is water).

Heat gained by ice: first it melts, then the melted water warms from $0\,^\circ\text{C}$0\,^ to $\theta$.
$Q_{gained}=m_{ice}L+m_{ice}c(\theta-0)=50\times336+50\times4.2\times\theta=16800+210\theta.$Q_gained=m_iceL+m_icec(-0)=50×336+50×4.2×=16800+210.

Heat lost by warm water: cools from $40\,^\circ\text{C}$40\,^ to $\theta$.
$Q_{lost}=m_wc(40-\theta)=200\times4.2\times(40-\theta)=840(40-\theta)=33600-840\theta.$Q_lost=m_wc(40-)=200×4.2×(40-)=840(40-)=33600-840.

Applying the principle of calorimetry ($Q_{gained}=Q_{lost}$Q_gained=Q_lost):
$16800+210\theta=33600-840\theta$16800+210=33600-840
$1050\theta=16800$1050=16800
$\theta=16\,^\circ\text{C}.$=16\,^.

The final temperature of the mixture is $16\,^\circ\text{C}$16\,^ (positive, so the assumption that all ice melts is valid).

Let the specific heat capacity of the metal be $c\text{ J g}^{-1}\,^\circ\text{C}^{-1}$c J g^-1\,^^-1. Final temperature $=30\,^\circ\text{C}$=30\,^.

Heat lost by hot metal (cools $100\to30$10030):
$Q_{lost}=200\times c\times(100-30)=200\times c\times70=14000c.$Q_lost=200× c×(100-30)=200× c×70=14000c.

Heat gained by water (warms $25\to30$2530):
$Q_{water}=250\times4.2\times(30-25)=250\times4.2\times5=5250\text{ J}.$Q_water=250×4.2×(30-25)=250×4.2×5=5250 J.

Heat gained by copper calorimeter (warms $25\to30$2530):
$Q_{cal}=150\times0.4\times(30-25)=150\times0.4\times5=300\text{ J}.$Q_cal=150×0.4×(30-25)=150×0.4×5=300 J.

Principle of calorimetry: heat lost by metal = heat gained by water + calorimeter.
$14000c=5250+300=5550$14000c=5250+300=5550
$c=\dfrac{5550}{14000}=0.396\approx0.40\text{ J g}^{-1}\,^\circ\text{C}^{-1}.$c=5550/14000=0.3960.40 J g^-1\,^^-1.

The specific heat capacity of the metal is about $0.40\text{ J g}^{-1}\,^\circ\text{C}^{-1}$0.40 J g^-1\,^^-1.

(i) The specific latent heat of fusion of a substance is the quantity of heat required to change unit mass of the substance from solid to liquid at its melting point, without any change in temperature. For ice it is $336\text{ J g}^{-1}$336 J g^-1.

(ii) During melting the temperature remains constant (at $0\,^\circ\text{C}$0\,^ for ice). The supplied heat is used as latent heat to change the state (break the bonds of the solid), not to raise the temperature.

(iii) $Q=mL=20\times336=6720\text{ J}.$Q=mL=20×336=6720 J.

(iv) Steam at $100\,^\circ\text{C}$100\,^ carries an extra $2260\text{ J g}^{-1}$2260 J g^-1 of latent heat of vaporisation. When it condenses on the skin it releases this large latent heat in addition to the heat given out while cooling, so it causes a far more severe burn than the same mass of water at $100\,^\circ\text{C}$100\,^.