If an average person jogs, hse produces $14.5 \times 10^3$ cal/min. This is removed by the evaporation of sweat. The amount of sweat evaporated per minute (assuming 1 kg requires $580 \times 10^3$ cal for evaparation) is (a) 0.025 kg (b) 2.25 kg (c) 0.05 kg (d) 0.20 kg

If an average person jogs, hse produces 14.5 × 10³ cal/min. This is removed by the evaporation of sweat. The a

NCERT Exemplar · Q2

14.5 \times 10^3

(a) 0.025 kg

(b) 2.25 kg

(d) 0.20 kg

Uttarakhand UbseMCQ· 1mImportance★★★★★est

29% · 10/35 Questions

✓ Free question

Concept understanding — Specific Heat Capacity

What is Specific Heat Capacity?

Imagine you have two identical stoves, two identical pots, and you put 1 kg of water in one pot and 1 kg of iron in the other. You turn both stoves to the same flame. After 2 minutes, the iron is scorching hot — you can't touch it. The water is still lukewarm.

Why? Because different substances need different amounts of heat to raise their temperature by the same amount. That's the core idea behind specific heat capacity.

The Intuition

Think of heat as "energy currency" and temperature rise as "buying a degree." Some materials are "cheap" — a little heat buys a big temperature rise. Others are "expensive" — you need to spend a lot of heat to get even a small rise.

Iron is cheap: a small heat input → large temperature jump.
Water is expensive: a large heat input → small temperature jump.

This "expensiveness" is what we call specific heat capacity. It tells you how much heat energy is needed to raise the temperature of 1 kg of a substance by 1 °C (or 1 K).

The Precise Definition

c = \frac{Q}{m \Delta T}

Where:

$c$ = specific heat capacity (J/kg·°C or J/kg·K)
$Q$ = heat energy supplied (J)
$m$ = mass of the substance (kg)
$\Delta T$ = change in temperature (°C or K)

In words: Specific heat capacity is the amount of heat required to raise the temperature of one kilogram of a substance by one degree Celsius (or one Kelvin).

Key Points to Remember

It's a property of the material, not the object. A small iron nail and a giant iron beam have the same $c$ value — but the beam needs more total heat because it has more mass.
Units matter. Common values:
- Water: $c = 4186 \text{ J/kg·°C}$ (or ≈ 4200 J/kg·°C in many problems)
- Iron: $c \approx 450 \text{ J/kg·°C}$
- Copper: $c \approx 390 \text{ J/kg·°C}$
Notice water's value is about 10 times that of iron — that's why water heats up so slowly compared to metals.
The formula works both ways. If a substance cools down, it releases the same amount of heat it would absorb to warm up by the same $\Delta T$ .

A Common Mistake to Avoid

Watch out

Don't confuse specific heat capacity ( $c$ ) with heat capacity ( $C$ ). Heat capacity is for an entire object: $C = mc$ . A large iron block can have a higher heat capacity than a small cup of water, even though iron's $c$ is much smaller. Always check: are we talking about per kg or for the whole thing?

Worked Example (Exam-Style)

Problem: How much heat is needed to raise the temperature of 2 kg of water from 20 °C to 50 °C? (Take $c_{\text{water}} = 4200 \text{ J/kg·°C}$ )

Solution:

$m = 2 \text{ kg}$
$\Delta T = 50 - 20 = 30 \text{ °C}$
$c = 4200 \text{ J/kg·°C}$

Q = mc\Delta T = 2 \times 4200 \times 30 = 252\,000 \text{ J} = 252 \text{ kJ}

Answer: 252 kJ of heat is required.

Why This Matters

Specific heat capacity explains countless everyday phenomena:

Why coastal cities have milder climates than inland ones (water's high $c$ stores heat)
Why a metal spoon in hot tea gets hot instantly while the tea stays hot longer
Why car radiators use water as coolant — it absorbs lots of heat without boiling

Important

Water has one of the highest specific heat capacities of any common substance. This single fact explains why oceans regulate Earth's climate, why your body uses water to maintain temperature, and why water is the go-to coolant in engines and power plants.

Final takeaway: Specific heat capacity is the "thermal inertia" of a material — how stubbornly it resists changing temperature when you add or remove heat. The higher the $c$ , the more heat you need to move its temperature.

A quick web search for "Specific Heat Capacity class 12 physics" or "Specific Heat Capacity class 11 physics" turns up this exact idea, because Specific Heat Capacity sits squarely within the Thermal Properties of Matter coverage of NCERT Class 11 Physics, so it is fair game for both CBSE board questions and competitive-exam numericals. Revisiting the NCERT Physics textbook exercises for this chapter alongside the walkthrough above is a solid way to convert this into exam-ready practice.

The heat produced by jogging is removed by the evaporation of sweat. By equating the heat produced per minute to the heat required for evaporation, we find that the amount of sweat evaporated per minute is $\boxed{0.025 \text{ kg}}$ .

When a person jogs, their body generates heat. To maintain a stable body temperature, this excess heat must be removed. One of the primary mechanisms for heat removal in humans is the evaporation of sweat from the skin. This process relies on the concept of latent heat of vaporization.

Latent heat of vaporization is the amount of heat energy required to change a substance from a liquid to a gaseous state at a constant temperature, without changing its temperature. For sweat (which is mostly water), a specific amount of heat is absorbed from the body for every kilogram of sweat that evaporates. This absorbed heat is what cools the body.

In this problem, we are given the rate at which heat is produced by jogging and the amount of heat required to evaporate a certain mass of sweat. We can equate the heat produced per minute to the heat absorbed by the evaporating sweat per minute to find the mass of sweat evaporated.

Here's how to solve the problem:

Identify the given quantities:
- Heat produced by jogging per minute ( $Q_{produced}$ ) = $14.5 \times 10^3$ cal/min
- Heat required to evaporate 1 kg of sweat ( $L_v$ ) = $580 \times 10^3$ cal/kg
State the principle of heat balance:

For the body to remove the heat produced, the heat generated must be equal to the heat absorbed by the evaporating sweat.

Q_{produced} = Q_{evaporated}

Formulate the equation for heat absorbed by evaporation: The heat absorbed by evaporation ( $Q_{evaporated}$ ) is given by the product of the mass of sweat evaporated ( $m$ ) and the latent heat of vaporization per unit mass ( $L_v$ ). $Q_{evaporated} = m \times L_v$

Q_{produced} = m \times L_v

Substitute the given values and solve for the mass of sweat evaporated ( $m$ ): We need to find the mass of sweat evaporated per minute. Let $m$ be this mass in kg.

14.5 \times 10^3 \text{ cal/min} = m \times (580 \times 10^3 \text{ cal/kg})

To find $m$, we rearrange the equation:

m = \frac{14.5 \times 10^3 \text{ cal/min}}{580 \times 10^3 \text{ cal/kg}}

The $10^3$ terms cancel out, and the 'cal' units cancel, leaving 'kg/min':

m = \frac{14.5}{580} \text{ kg/min}

Now, perform the division:

m = \frac{145}{5800} \text{ kg/min}

m = \frac{29}{1160} \text{ kg/min}

Since $1160 = 29 \times 40$:

m = \frac{1}{40} \text{ kg/min}

m = 0.025 \text{ kg/min}

✓Final answer

\boxed{0.025 \text{ kg}}

←Previous questionQ1 — An ideal gas undergoes four different processes starting from the same…→Next questionQ3 — An ideal gas is taken from state 1 to state 2 along a curve on a P-V d…

Q.If an average person jogs, hse produces 14.5×10314.5 \times 10^314.5×103 cal/min. This is removed by the evaporation of sweat. The amount of sweat evaporated per minute (assuming 1 kg requires 580×103580 \times 10^3580×103 cal for evaparation) is

🎓Unlock everything free for 14 days

Q.If an average person jogs, hse produces $14.5 \times 10^3$ cal/min. This is removed by the evaporation of sweat. The amount of sweat evaporated per minute (assuming 1 kg requires $580 \times 10^3$ cal for evaparation) is

Unlock everything free for 14 days