Calorimetry — Measuring Heat Transfer

Heat, Temperature, and Why They’re Different

Before we measure heat, we need to know what it is. Temperature is a measure of the average kinetic energy of molecules — it tells you how “hot” something is. Heat is energy in transit from a hotter body to a cooler one.

A swimming pool at 30°C and a cup of tea at 70°C: the tea is hotter (higher temperature), but the pool contains far more thermal energy (more molecules). When you pour the tea into the pool, heat flows from the tea to the pool — from higher temperature to lower temperature.

Calorimetry is the science of measuring this heat transfer. We use a device called a calorimeter — essentially a well-insulated container that minimizes heat exchange with the surroundings so we can measure only what we’re interested in.

Specific Heat Capacity

Q = mc\Delta T

$Q$ = heat absorbed or released (in Joules)
$m$ = mass of the substance (in kg or g)
$c$ = specific heat capacity (J/kg·K or J/g·°C)
$\Delta T = T_f - T_i$ = change in temperature

Specific heat capacity $c$ is the heat required to raise 1 kg of a substance by 1 K (or 1°C). It’s a property of the material itself.

Key values to memorize:

Substance	$c$ (J/g·°C)
Water	4.18
Ice	2.09
Steam	2.01
Copper	0.39
Aluminium	0.90
Iron	0.45

Water has the highest specific heat of common substances. This is why coastal cities have milder climates and why water is used as a coolant in car engines.

When $\Delta T$ is the same for different substances of the same mass, the one with higher $c$ absorbs more heat. Water “resists” temperature change much more than metals — so metals heat up and cool down quickly, while water changes temperature slowly.

Latent Heat — Phase Changes Without Temperature Change

When a substance changes phase (solid → liquid, liquid → gas), it absorbs or releases heat WITHOUT any temperature change. This heat is called latent heat.

Q = mL

$L$ = specific latent heat (J/kg or J/g)
No $\Delta T$ term because temperature doesn’t change during a phase change

Latent heat of fusion ( $L_f$ ): Heat needed to convert 1 kg of solid to liquid at the melting point. For ice: $L_f = 334$ J/g $= 80$ cal/g

Latent heat of vaporization ( $L_v$ ): Heat needed to convert 1 kg of liquid to vapour at the boiling point. For water: $L_v = 2260$ J/g $= 540$ cal/g

The latent heat of vaporization of water (2260 J/g) is about 7 times larger than the latent heat of fusion (334 J/g). This is why steam burns are more severe than boiling water burns — steam releases extra heat as it condenses on your skin.

Heating Curve of Water

The heating curve shows what happens as we supply heat uniformly to ice at $-20°C$ :

Ice warms from $-20°C$ to $0°C$ : $Q = mc_{ice}\Delta T$ , temperature rises
Ice melts at $0°C$ : $Q = mL_f$ , temperature stays at $0°C$
Water warms from $0°C$ to $100°C$ : $Q = mc_{water}\Delta T$ , temperature rises
Water boils at $100°C$ : $Q = mL_v$ , temperature stays at $100°C$
Steam warms above $100°C$ : $Q = mc_{steam}\Delta T$ , temperature rises

The flat portions of the curve (at $0°C$ and $100°C$ ) represent latent heat being absorbed at constant temperature.

The Principle of Calorimetry

When two substances at different temperatures are mixed in a calorimeter (assuming no heat loss to surroundings):

\text{Heat lost by hot body} = \text{Heat gained by cold body}

m_1 c_1 (T_1 - T_{mix}) = m_2 c_2 (T_{mix} - T_2)

where $T_1 > T_{mix} > T_2$

This assumes the calorimeter itself absorbs negligible heat (or we account for the water equivalent of the calorimeter).

Water equivalent: The mass of water that would absorb the same heat as the calorimeter for the same temperature rise. If the calorimeter has mass $m_c$ and specific heat $c_c$ : Water equivalent $= m_c c_c / c_{water}$ .

Solved Examples

Example 1 — CBSE Level

100 g of water at 80°C is mixed with 200 g of water at 20°C. Find the final temperature. ( $c_{water} = 4.18$ J/g·°C)

Heat lost by hot water = Heat gained by cold water

100 \times 4.18 \times (80 - T) = 200 \times 4.18 \times (T - 20)

100(80 - T) = 200(T - 20)

8000 - 100T = 200T - 4000

12000 = 300T

T = 40°C

When both substances have the same specific heat (like water + water), the $c$ cancels out. The final temperature is simply the mass-weighted average: $T = \frac{m_1 T_1 + m_2 T_2}{m_1 + m_2} = \frac{100 \times 80 + 200 \times 20}{300} = \frac{12000}{300} = 40°C$ .

Example 2 — JEE Main Level

50 g of ice at 0°C is mixed with 200 g of water at 40°C. Find the final temperature. ( $L_f = 80$ cal/g, $c_w = 1$ cal/g·°C)

First, check if all ice melts. Heat available from hot water cooling to 0°C:

Q_{available} = 200 \times 1 \times 40 = 8000 \text{ cal}

Heat needed to melt all ice:

Q_{melt} = 50 \times 80 = 4000 \text{ cal}

Since $8000 > 4000$ , all ice melts. Remaining heat: $8000 - 4000 = 4000$ cal.

Now, 250 g of water at 0°C absorbs this heat:

4000 = 250 \times 1 \times T

T = 16°C

Final temperature = 16°C

Example 3 — JEE Advanced Level

A 100 g copper block at 200°C is dropped into 300 g water at 30°C in a copper calorimeter of mass 100 g. Find the final temperature. ( $c_{Cu} = 0.1$ cal/g·°C, $c_w = 1$ cal/g·°C)

Heat lost by copper block:

Q_1 = 100 \times 0.1 \times (200 - T)

Heat gained by water + calorimeter:

Q_2 = 300 \times 1 \times (T - 30) + 100 \times 0.1 \times (T - 30) = (300 + 10)(T - 30) = 310(T - 30)

Setting $Q_1 = Q_2$ :

10(200 - T) = 310(T - 30)

2000 - 10T = 310T - 9300

11300 = 320T

T = 35.3°C

Common Mistakes to Avoid

Mistake 1: Using a single formula $Q = mc\Delta T$ when a phase change is involved. Always check if the system crosses 0°C or 100°C, and account for latent heat separately.

Mistake 2: Forgetting to check whether all the ice/solid actually melts. If heat available is less than heat needed for complete melting, the final temperature is the melting point (e.g., 0°C for ice), not some value above it.

Mistake 3: Getting the sign wrong. Heat lost by the hot body is positive (temperature decreases), heat gained by cold body is positive (temperature increases). Set them equal without worrying about signs if you write the equation as: $Q_{lost} = Q_{gained}$ .

Mistake 4: Using inconsistent units. If mass is in grams, use $c$ in J/g·°C. If mass is in kg, use $c$ in J/kg·K. Mixing units is the most common source of errors.

Mistake 5: Ignoring the calorimeter’s heat capacity in advanced problems. If the calorimeter is made of metal and has a given mass and specific heat, you must include it in the “heat gained” side.

Exam-Specific Tips

CBSE Class 10: Focus on the principle of calorimetry and its application. Questions typically give two substances mixing and ask for the final temperature. Know the conversion: 1 cal = 4.18 J.

CBSE Class 11: Latent heat problems with phase changes are important. The heating curve of water is a common diagram question.

JEE Main: Ice-water mixture problems appear every other year. Always check if full phase change occurs before writing equations. Also know the concept of water equivalent.

NEET: Calorimetry is straightforward in NEET — mostly numerical problems using $Q = mc\Delta T$ and $Q = mL$ are sufficient.

Practice Questions

Q1. How much heat is needed to convert 50 g of ice at $-10°C$ to water at $50°C$ ? ( $c_{ice} = 2.09$ J/g·°C, $L_f = 334$ J/g, $c_w = 4.18$ J/g·°C)

Heat to warm ice: $Q_1 = 50 \times 2.09 \times 10 = 1045$ J

Heat to melt ice: $Q_2 = 50 \times 334 = 16700$ J

Heat to warm water: $Q_3 = 50 \times 4.18 \times 50 = 10450$ J

Total = $1045 + 16700 + 10450 = 28195$ J $\approx 28.2$ kJ

Q2. 100 g of steam at 100°C is passed into 500 g of water at 20°C. Find the final temperature. ( $L_v = 540$ cal/g)

Heat available from condensation alone: $100 \times 540 = 54000$ cal.

Heat needed to warm 500 g water from 20°C to 100°C: $500 \times 1 \times 80 = 40000$ cal.

Since $54000 > 40000$ , not all steam is needed. After water reaches 100°C, steam starts condensing. Let $x$ g of steam condense: $x \times 540 = 40000 \Rightarrow x = 74.1$ g. Remaining steam = 25.9 g. But we must also warm condensed steam to check…

Actually, total water at 100°C after all steam condenses: $100 \times 540 + 100 \times 1 \times 0 = 54000$ cal released. Water absorbs: $500 \times 80 = 40000$ cal to reach 100°C. Extra heat: $54000 - 40000 = 14000$ cal. But 600 g of water at 100°C can’t rise above 100°C. Final temperature = 100°C (some steam may remain; there is excess heat, meaning all water reaches 100°C and final temperature is 100°C).

Q3. 200 g of a metal at 100°C is placed in 400 g of water at 25°C. The final temperature is 30°C. Find the specific heat of the metal.

Heat lost by metal = Heat gained by water

$200 \times c \times (100 - 30) = 400 \times 4.18 \times (30 - 25)$

$200 \times c \times 70 = 400 \times 4.18 \times 5$

$14000c = 8360$

$c = 0.597$ J/g·°C $\approx 0.6$ J/g·°C

FAQs

Q: Why does sweating cool the body?

Sweat (water) on the skin evaporates, absorbing the latent heat of vaporization (2260 J/g) from your skin. This large heat removal cools the body effectively. The same principle explains why an earthen pot keeps water cool.

Q: Why does ice at 0°C feel colder than water at 0°C?

When ice at 0°C touches your skin, it first absorbs heat to melt (latent heat = 334 J/g) and then the resulting cold water warms up. Water at 0°C only needs to warm up. So ice extracts more total heat from your skin.

Q: What is the calorie and how does it relate to the food calorie?

1 calorie = 4.18 J (heat to raise 1 g of water by 1°C). 1 food Calorie (capital C) = 1 kcal = 4180 J. So a 400 Calorie snack contains 1.672 × 10⁶ J of chemical energy.

Q: Why is $L_v$ so much larger than $L_f$ ?

During melting, molecules only need to break some intermolecular bonds (they still remain close together as liquid). During vaporization, molecules must break most intermolecular bonds and expand against atmospheric pressure to move far apart. Both of these require much more energy.