Class 10 Physics • Chapter 11 (Deep Detail)
Heat Capacity ($C'$): Amount of heat required to raise the temperature of the whole body by 1°C (or 1 K).
$C' = m \times c$. Unit: J/K.
Amount of heat required to raise the temperature of unit mass (1 kg) of a substance by 1°C.
$$ Q = m c \Delta T $$Unit: J kg$^{-1}$ K$^{-1}$.
Note: Water has very high specific heat capacity ($4200 \text{ J kg}^{-1} \text{ K}^{-1}$).
NUMERICAL Calculate the heat energy required to raise the temperature of 2 kg of water from $10^\circ C$ to $50^\circ C$. ($c = 4200 \text{ J/kgK}$).
Solution:
$Q = mc \Delta T = 2 \times 4200 \times (50 - 10)$.
$Q = 8400 \times 40 = 336,000 \text{ J} = 336 \text{ kJ}$.
NUMERICAL A body of mass 1.5 kg requires 4500 J of heat to raise its temperature by $10^\circ C$. Calculate its specific heat capacity.
Solution:
$c = \frac{Q}{m \Delta T} = \frac{4500}{1.5 \times 10}$
$= \frac{4500}{15} = 300 \text{ J/kgK}$.
Q: Why is water used as a coolant in car radiators?
Ans: High specific heat capacity ($4200 \text{ J/kgK}$). Can absorb large heat with small temp rise.
CONCEPTUAL Distinguish between Heat Capacity and Specific Heat Capacity.
Ans: Heat Capacity is for the whole body ($C' = mc$). Specific Heat Capacity is for unit mass ($c = C'/m$).
If no heat is lost to the surroundings:
Note: This principle is also known as the Method of Mixtures.
NUMERICAL 100 g of hot water at $80^\circ C$ is added to 200 g of cold water at $10^\circ C$. Find final temperature. (Neglect vessel heat).
Solution:
Let final temp be $T$.
Heat lost = $100 \times c \times (80 - T)$. Heat gained = $200 \times c \times (T - 10)$.
$100 (80-T) = 200 (T-10)$
$\implies 80 - T = 2T - 20$.
$3T = 100$
$\implies T = 33.33^\circ C$.
Q: What assumption is made in the principle of mixtures?
Ans: No heat is lost to the surroundings or the container.
NUMERICAL A copper vessel of mass 100g contains 150g of water at 50°C. How much ice at 0°C is needed to cool it to 5°C? (Given $c_{Cu} = 0.4 \text{ J/gK}$, $c_{w} = 4.2 \text{ J/gK}$, $L_{ice} = 336 \text{ J/g}$).
Solution:
Heat Lost by (Water + Vessel) = $(150 \times 4.2 + 100 \times 0.4) \times (50-5)$
$= 670 \times 45 = 30150 \text{ J}$.
Heat Gained by Ice = $m \times 336 + m \times 4.2 \times (5-0) = 357m$.
$357m = 30150 \implies m = 84.45 \text{ g}$.
NUMERICAL 40 g of water at 60°C is poured into a vessel containing 50 g of water at 20°C. Calculate the final temperature. (Neglect vessel).
Solution:
$40c(60-T) = 50c(T-20)$
$\implies 4(60-T) = 5(T-20)$
$\implies 240-4T = 5T-100$
$\implies 9T=340$
$\implies T=37.77^\circ C$.
Heat energy absorbed or released at constant temperature during change of state.
Latent Heat of Fusion ($L_f$): Solid $\leftrightarrow$ Liquid at Melting Point. ($Q = m L_f$). For Ice: $L_f = 336 \text{ J/g} = 336000 \text{ J/kg}$.
Latent Heat of Vaporisation ($L_v$): Liquid $\leftrightarrow$ Gas at Boiling Point. ($Q = m L_v$). For Steam: $L_v = 2260 \text{ J/g}$.
NUMERICAL Calculate heat required to melt 500g of ice at $0^\circ C$ to water at $0^\circ C$. ($L_f = 336 \text{ J/g}$).
Solution:
$Q = m L_f = 500 \times 336$.
$Q = 168,000 \text{ J} = 168 \text{ kJ}$.
GRAPH ANALYSIS In the heating curve of ice, why does the temperature remain constant during melting (Region 2)?
Ans: The heat supplied during this phase is used as Latent Heat to break the intermolecular bonds and change the state from solid to liquid, rather than increasing the kinetic energy (temperature) of the molecules.
NUMERICAL Convert 100 g ice at -10°C to water at 10°C.
Solution: 1. Ice (-10 to 0): $0.1 \times 2100 \times 10 = 2100 J$. 2. Melt: $0.1 \times 336000 = 33600 J$. 3. Water (0 to 10): $0.1 \times 4200 \times 10 = 4200 J$. Total = $39900 \text{ J}$.
REASONING Why do steam burns feel more severe than boiling water burns?
Ans: Steam contains extra latent heat of vaporization ($2260 \text{ J/g}$).
Remember: $1 \text{ calorie} = 4.2 \text{ Joules}$.
Sea Breeze & Land Breeze: Due to high specific heat capacity, water warms up and cools down slower than land. This creates pressure differences causing breezes.
NUMERICAL 40 g of water at 60°C is poured into a vessel containing 50 g of water at 20°C. Calculate the final temperature of the mixture. (Neglect heat capacity of vessel).
Solution:
Let final temp be $T$.
Heat Lost by Hot Water = $m_1 c (60 - T) = 40 \times c \times (60 - T)$.
Heat Gained by Cold Water = $m_2 c (T - 20) = 50 \times c \times (T - 20)$.
$40c(60-T) = 50c(T-20)$.
$\implies 4(60-T) = 5(T-20)$.
$240 - 4T = 5T - 100$
$\implies 9T = 340$
$\implies T = 37.77^\circ \text{C}$.
Ans: 37.77°C.
NUMERICAL Calculate the amount of heat energy required to convert 100 g of ice at $-10^\circ \text{C}$ into water at $10^\circ \text{C}$. (Given $c_{ice} = 2100 \text{ J/kgK}$, $c_{water} = 4200 \text{ J/kgK}$, $L_f = 336000 \text{ J/kg}$).
Solution:
Mass $m = 0.1 \text{ kg}$.
1. Ice (-10 to 0): $Q_1 = 0.1 \times 2100 \times 10 = 2100 \text{ J}$.
2. Melting: $Q_2 = 0.1 \times 336000 = 33600 \text{ J}$.
3. Water (0 to 10): $Q_3 = 0.1 \times 4200 \times 10 = 4200 \text{ J}$.
Total Heat $Q = 2100 + 33600 + 4200 = 39900 \text{ J}$.
Ans: 39,900 J.
REASONING Why do burns caused by steam at 100°C appear more severe than those caused by boiling water at 100°C?
Solution:
Steam at 100°C contains Latent Heat of Vaporization ($2260 \text{ J/g}$) in addition to the heat energy possessed by boiling water at the same temperature. When steam condenses on skin, it releases this massive extra heat, causing more severe burns.