Q.A geyser heats water flowing at the rate of $3.0$ litres per minute from $27\,^{\circ}\text{C}$ to $77\,^{\circ}\text{C}$. If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is $4.0 \times 10^{4}\ \text{J/g}$?

Concept and Intuition

When a hot object is placed on ice, the object cools down by transferring heat to the ice. The ice uses that heat to change phase from solid to liquid (melting) at a constant temperature of 0°C. The key principle is conservation of energy: the heat lost by the block equals the heat gained by the ice (assuming no heat loss to the surroundings). The block’s heat loss depends on its mass, specific heat, and temperature change. The ice’s heat gain depends on the mass of ice melted and the latent heat of fusion. By equating these, we can solve for the unknown mass of the block.

Step-by-step solution

1. Identify the heat lost by the block The block cools from $500^\circ \mathrm{C}$ to $0^\circ \mathrm{C}$ (the melting point of ice). The heat lost is given by:

where $c = 0.39 \, \mathrm{J\,g^{-1}\,^\circ C^{-1}}$ and $\Delta T = 500 - 0 = 500^\circ \mathrm{C}$.

So:

2. Identify the heat gained by the ice The ice melts at $0^\circ \mathrm{C}$ using the latent heat of fusion. The heat gained is:

where $m_{\text{ice}} = 1.455 \, \mathrm{kg} = 1455 \, \mathrm{g}$ (since specific heat is given in $\mathrm{J\,g^{-1}}$, we work in grams) and $L_f = 335 \, \mathrm{J\,g^{-1}}$.

So:

3. Apply conservation of energy Assuming no heat is lost to the environment, the heat lost by the block equals the heat gained by the ice:

4. Solve for $m_{\text{block}}$ First, compute the right-hand side:

(Alternatively, $1455 \times 335 = 487425$).

The left-hand side simplifies:

So the equation becomes:

Divide both sides by 195:

Perform the division:

So:

But more precisely:

However, note that $195 \times 2500 = 487500$ and $487500 - 487425 = 75$, so the exact value is $2500 - 75/195 = 2500 - 5/13$. But wait—check if the numbers are exact:

$1455 \times 335 = 1455 \times 335$. Let’s compute exactly:

$1455 \times 335 = 1455 \times (300 + 35) = 436500 + 50925 = 487425$.

Now $487425 / 195$: divide numerator and denominator by 15:

$195 \div 15 = 13$, $487425 \div 15 = 32495$. So $m_{\text{block}} = 32495 / 13$.

$13 \times 2500 = 32500$, so $32495 = 13 \times 2500 - 5$, thus $m_{\text{block}} = 2500 - 5/13 \approx 2499.615 \, \mathrm{g}$.

That is approximately $2.4996 \, \mathrm{kg}$, which rounds to $2.5 \, \mathrm{kg}$.

5. Match with the options The options are: (A) 0.67 kg, (B) 2.5 kg, (C) 1.455 kg, (D) 2.67 kg. Our result is 2.5 kg, which matches option (B).

The problem is a classic trap: it says “having the same heat capacities.” Many students read that as “same specific heat,” but heat capacity is the product of mass and specific heat: $C_{\text{body}} = m c$. So “same heat capacities” means

where $C_1$ and $C_2$ are the specific heats (per unit mass). The composite body has total mass $m_1 + m_2$ and total heat capacity $m_1 C_1 + m_2 C_2$. Its specific heat is that total divided by total mass.

1. Set up the equality of heat capacities Let the two bodies have masses $m_1$ and $m_2$ and specific heats $C_1$ and $C_2$. Their heat capacities are equal:

This is the only relation between the masses.

2. Express masses in terms of one another From the equality,

3. Total heat capacity of the composite The composite’s total heat capacity is the sum:

Substitute $m_2$:

4. Total mass of the composite

5. Specific heat of the composite Specific heat = total heat capacity ÷ total mass:

Convert masses to kg to match the given specific heat unit (J kg$^{-1}\,^\circ$C$^{-1}$): iron $0.200$ kg, water $0.230$ kg, calorimeter water-equivalent $0.020$ kg (total effective water $0.250$ kg). Take water $c_w = 4200\text{ J kg}^{-1}\,^\circ\text{C}^{-1}$ and iron $c = 470$.

1. Heat gained by water + calorimeter ($20\to 60\,^\circ$C):

2. Heat lost by iron ($T \to 60\,^\circ$C):

3. Energy balance $Q_{\text{lost}} = Q_{\text{gain}}$:

Note: uses the standard specific heat of water $4200\text{ J kg}^{-1}\,^\circ\text{C}^{-1}$; failing to convert grams to kilograms is the trap that produces the spurious $\sim\!953\,^\circ$C figure.

Concept & Intuition

When the ice falls, its gravitational potential energy is converted into kinetic energy, and upon impact, most of that kinetic energy is dissipated as heat. The problem tells us that 60% of the total mechanical energy becomes thermal energy absorbed by the ice. That heat does two things: it raises the temperature of the entire 1 kg block from –1.5°C to 0°C (sensible heat), and the remaining heat then melts a portion of the ice (latent heat). We need to find how much ice actually melts.

1. Calculate the total gravitational potential energy lost The ice has mass $m = 1\ \mathrm{kg} = 1000\ \mathrm{g}$. Height $h = 1.5\ \mathrm{km} = 1500\ \mathrm{m}$. Gravitational potential energy:

2. Convert this energy into calories Since the specific heat and latent heat are given in calories, we work in calories. $1\ \mathrm{cal} = 4.184\ \mathrm{J}$, so

3. Only 60% of this energy becomes heat

4. First, warm the ice from –1.5°C to 0°C Specific heat of ice $c = 0.5\ \mathrm{cal\,g^{-1}\,°C^{-1}}$. Mass of ice = 1000 g. Heat needed to raise temperature:

5. Remaining heat goes into melting

6. Find the mass of ice melted Latent heat of fusion $L = 80\ \mathrm{cal/g}$.

Rounding gives about 16.9 g.

Step 1 — The law and why this form works.

Newton's law of cooling states that the rate of heat loss is proportional to the excess of the body's temperature over its surroundings:

When the temperature drop over the interval is small compared with the excess temperature, we may replace $\theta$ by the average temperature over the interval:

Step 2 — Calibrate $k$ from the first cooling.

$\theta_1 = 65.5^\circ$C, $\theta_2 = 62.5^\circ$C, $t = 1$ min, $\theta_0 = 22.5^\circ$C.

Step 3 — Apply to the second cooling.

$\theta_1 = 46.5^\circ$C, $\theta_2 = 40.5^\circ$C, unknown time $t$.

Step 4 — Solve for $t$.

Physically this makes sense: the tea is now much closer to room temperature, so it cools more slowly — a 6°C fall takes about 4 minutes, whereas a 3°C fall took only 1 minute when the tea was hot.

ANSWER: A