Calorimetry — ICSE Class 10 Physics Important Questions

Answer: (b) $\text{J kg}^{-1}\,^\circ\text{C}^{-1}$J kg^-1 ° C^-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·° C=J kg^-1 ° C^-1. The unit $\text{J }^\circ\text{C}^{-1}$J ° C^-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 ° C 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 ° C^-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/mc∝ 1/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 ° C (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 ° C^-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.

Step 1, melting at $0\,^\circ\text{C}$0 ° C: $Q_1=mL=2\times336000=672000\text{ J}$Q₁=mL=2×336000=672000 J.

Step 2, warming the water from $0$0 to $30\,^\circ\text{C}$30 ° C: $Q_2=mc\,\Delta T=2\times4200\times30=252000\text{ J}$Q₂=mc Δ T=2×4200×30=252000 J.

Total: $Q=672000+252000=924000\text{ J}=9.24\times10^5\text{ J}$Q=672000+252000=924000 J=9.24×10⁵ J.

Let the final temperature be $\theta$θ. 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-θ).

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

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

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

(a) Heat capacity of a body is the heat needed to raise its temperature by $1\,^\circ\text{C}$1 ° C; SI unit $\text{J }^\circ\text{C}^{-1}$J ° C^-1 (or $\text{J K}^{-1}$J K^-1). Heat capacity $=$= mass $\times$× specific heat capacity: $C'=mc$C'=mc.

(b) $C'=0.5\times400=200\,\text{J }^\circ\text{C}^{-1}$C'=0.5×400=200 J ° C^-1. Heat needed $=C'\times\Delta T=200\times10=2000\text{ J}$=C'× Δ T=200×10=2000 J.

Let the final temperature be $\theta\,^\circ\text{C}$θ ° 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 ° C to $\theta$θ.
$Q_{gained}=m_{ice}L+m_{ice}c(\theta-0)=50\times336+50\times4.2\times\theta=16800+210\theta.$Q_gained=m_iceL+m_icec(θ-0)=50×336+50×4.2× θ=16800+210θ.

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

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

The final temperature of the mixture is $16\,^\circ\text{C}$16 ° C (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 ° C^-1. Final temperature $=30\,^\circ\text{C}$=30 ° C.

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

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

Heat gained by copper calorimeter (warms $25\to30$25→30):
$Q_{cal}=150\times0.4\times(30-25)=150\times0.4\times5=300\text{ J}.$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=0.396≈0.40 J g^-1 ° C^-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 ° C^-1.

(i) AB: the ice warms from $-10\,^\circ\text{C}$-10 ° C to $0\,^\circ\text{C}$0 ° C. BC: the ice melts at $0\,^\circ\text{C}$0 ° C. CD: the water warms from $0\,^\circ\text{C}$0 ° C to $40\,^\circ\text{C}$40 ° C.

(ii) During melting the heat supplied is used as latent heat to change the solid into liquid, not to raise the temperature.

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

(iv) Each gram of ice first takes $336\text{ J}$336 J from the drink just to melt, and only then warms like water. So ice absorbs much more heat from the drink than water at the same temperature.