Temperature and temperature difference

An electric kettle heats water from the tap temperature of $20\ \celsius$ to the $70\ \celsius$ needed by grandma for her herbal tea. Calculate the temperature change $\Delta T$ of the water.

A pot with $2\ \text{kg}$ of water at $90\ \celsius$ is removed from the stove and loses $167\,200\ \text{J}$ to the kitchen surroundings.

1. a) Calculate how many degrees the temperature of the water drops.
2. b) What is the final temperature of the water?

You leave a cup of coffee at $80\ \celsius$ on a kitchen table that is at $20\ \celsius$, as shown in the figure.

1. a) In which direction does heat flow: from the coffee to the air or from the air to the coffee?
2. b) What temperature does the coffee tend towards if you leave it for a long time?

Aitana leaves the spoon inside the pot of hot soup while she sets the table. When she returns, the spoon is as hot as the soup.

1. a) Explain in your own words what thermal equilibrium is using this example.
2. b) When two bodies are in thermal equilibrium, does heat continue to pass from one to the other?

The heat that must be supplied to a mass $m$ of a substance to raise its temperature by $\Delta T$ is calculated using $Q = m \times c \times \Delta T$, where $c$ is the specific heat capacity. For water, $c = 4180\ \text{J}/(\text{kg}\times\celsius)$. Calculate the heat required to heat $1\ \text{kg}$ of water so that its temperature rises by $10\ \celsius$.

You want to heat water for pasta and are considering a saucepan with $1\ \text{kg}$ of water and a pot with $2\ \text{kg}$, raising the temperature by the same amount in both cases.

1. a) Which of the two requires more heat?
2. b) Justify your answer using the formula $Q = m \times c \times \Delta T$.

In the formula $Q = m \times c \times \Delta T$, indicate what each letter represents, which quantity it measures, and what unit it is expressed in using the International System of Units.

The specific heat capacity of water is $c = 4180\ \text{J}/(\text{kg}\times\celsius)$. Explain in your own words what this number means: how many joules are required and to achieve what, exactly?

State whether the statement is true or false and correct the idea: "The greater the temperature rise I want to achieve in a saucepan of water, the less heat I need to supply to it."

In the laboratory, you heat $500\ \text{mL}$ of water from $20\ \celsius$ to $60\ \celsius$. Remember that $1\ \text{L}$ of water has a mass of $1\ \text{kg}$. The specific heat of water is $c_e = 4184\ \text{J}/(\text{kg}\times\celsius)$.

1. a) Express the mass of the water in kilograms.
2. b) Calculate the heat absorbed by the water.

A domestic electric water heater stores $150\ \text{L}$ of water (about $150\ \text{kg}$) and heats it from $15\ \celsius$ to $55\ \celsius$.

1. a) Calculate the total heat required by the water in the heater.
2. b) Express the result in standard form.

An inflatable garden pool contains $2000\ \text{kg}$ of water. On a sunny day, the water temperature only manages to rise by $1\ \celsius$.

1. a) Calculate the energy that the water had to absorb to rise by that single degree and express it in standard form.
2. b) Compare this result with the $41\,800\ \text{J}$ required to raise one litre of kitchen water by $10\ \celsius$.
3. c) Use the comparison to explain why pools and the sea heat up so slowly even though the Sun shines all day.

A pot with $2\ \text{kg}$ of water is heated from $20\ \celsius$ to $70\ \celsius$. Remember that for water $c = 4180\ \text{J}/(\text{kg}\times\celsius)$.

1. a) Calculate the temperature change.
2. b) Calculate the heat received by the water.

A heater supplies $83\,600\ \text{J}$ to a certain mass of water, and its temperature rises by $20\ \celsius$.

1. a) Rearrange the formula $Q = m \times c \times \Delta T$ to make mass the subject.
2. b) Calculate how many kilograms of water have been heated.

In a frying pan there is $1\ \text{kg}$ of oil ($c = 2000\ \text{J}/(\text{kg}\times\celsius)$) and in a saucepan, $1\ \text{kg}$ of water ($c = 4180\ \text{J}/(\text{kg}\times\celsius)$). Both receive the same amount of heat: $41\,800\ \text{J}$, as shown in the figure.

1. a) Calculate the temperature rise for each.
2. b) Explain using these numbers why the oil in the frying pan heats up much faster than the water in the saucepan.

In a saucepan, you mix $1\ \text{L}$ of water at $80\ \celsius$ with $1\ \text{L}$ of water at $20\ \celsius$, as shown in the figure.

1. a) Calculate the final temperature of the mixture, assuming there are no heat losses.
2. b) Explain why, since there are equal masses of the same substance, the final temperature is exactly the average of the two.

In an insulated flask, you mix $2\ \text{kg}$ of water at $60\ \celsius$ with $1\ \text{kg}$ of water at $30\ \celsius$. The heat released by the hot water is equal to the heat gained by the cold water.

1. a) Set up the energy balance and calculate the final temperature.
2. b) Explain why the result is not $45\ \celsius$, the simple average between $60\ \celsius$ and $30\ \celsius$.

In an insulated flask, you mix $3\ \text{kg}$ of water at $80\ \celsius$ with $1\ \text{kg}$ of water at $20\ \celsius$.

1. a) Set up the balance $Q_{\text{cedido}} = Q_{\text{ganado}}$ using the formula $Q = m \times c \times \Delta T$.
2. b) Notice that the specific heat capacity $c$ appears on both sides and cancels out: explain why it is not necessary to know it when both masses are of the same substance.
3. c) Calculate the final temperature of the mixture.

You have $2\ \text{kg}$ of very hot water at $80\ \celsius$, and you want to bring it to $50\ \celsius$ by adding cold tap water, which comes out at $20\ \celsius$, as shown in the figure.

1. a) Set up the heat balance by calling the mass of cold water to be added $m$.
2. b) Calculate this mass.
3. c) Check the result by verifying that the heat released by the hot water equals the heat gained by the cold water.

In a gym shower, the hot water comes out at $70\ \celsius$ and the cold water at $20\ \celsius$. You want to fill a $50\ \text{L}$ bucket with water at $40\ \celsius$ to mop the floor, as shown in the figure.

1. a) Set up the heat balance and find the proportion in which you must mix hot and cold water.
2. b) Calculate how many litres from each tap you need for the $50\ \text{L}$.
3. c) Check that the mixture of these litres is indeed at $40\ \celsius$.

To cool down a tea, you pour $0.1\ \text{kg}$ of cold water at $10\ \celsius$ into the cup, which contains $0.3\ \text{kg}$ of tea (practically water) at $80\ \celsius$.

1. a) Set up the heat balance using the data in kilograms.
2. b) Calculate the final temperature of the mixture.
3. c) If it still seems too hot to you, would you add more cold water or wait for the cup to lose heat to the air? Justify which will cool it down faster and why.

You have filled the bathtub with $100\ \text{L}$ of water at $50\ \celsius$, which is too hot for a bath. You turn on the cold water tap, which comes out at $10\ \celsius$, and want to leave the bath at $30\ \celsius$.

1. a) Set up the heat balance between the water in the bathtub and the cold water you add.
2. b) Calculate how many litres of cold water you need.
3. c) What will be the total volume of water in the bathtub at the end?

In the laboratory, you mix equal masses of water at $80\ \celsius$ and $20\ \celsius$ inside a beaker. The ideal calculation predicts a final temperature of $50\ \celsius$, but the thermometer reads $47\ \celsius$.

1. a) Explain to which bodies, apart from the cold water, the hot water has transferred heat in reality.
2. b) Does this result contradict the first law? Justify your answer by indicating where the "missing" energy is.

A classmate claims: "If I mix a lot of cold water with a little hot water, the mixture can end up hotter than the hot water was at the beginning, because there is much more water."

1. a) Reason, considering which body releases heat and which gains it, between which two values the final temperature of a mixture of two masses of water must always lie.
2. b) Explain then whether the classmate's statement is possible.
3. c) What condition would cause the final temperature to be very close to that of the hot water?

Unit challenge. How much energy is lost when heating water at home? Assess the thermal efficiency of your kitchen by connecting electrical work, heat, and temperature.

1. Data collection: Look up the rated electrical input power $P$ (not the useful output power that some microwaves state) of your microwave or kettle on the appliance's technical label. Measure the initial temperature $T_0$ of the water before switching on the appliance. Heat $m = 0.25\text{ kg}$ of water for four different times (for example, $15$, $30$, $45$, and $60\text{ s}$; shorten them if your appliance is very powerful, such as a kettle), without letting the water exceed about $80\text{ }^\circ\text{C}$, since once it boils $Q = m \times c \times \Delta T$ is no longer valid, recording the values with a stopwatch ($t \pm 0.5\text{ s}$) and a kitchen thermometer ($T \pm 0.5\text{ }^\circ\text{C}$).

2. Processing: Create a table with $t$, $T$, and the temperature difference $\Delta T = T - T_0$. Plot $T$ versus $t$. Calculate the electrical work supplied ($W = P \times t$) and the useful heat absorbed by the water ($Q = m \times c \times \Delta T$, with $c = 4184\text{ J/(kg}\times^\circ\text{C)}$).

3. Report and proposal: Compare $W$ and $Q$, determine the energy efficiency of the process using $\eta = (Q / W) \times 100$, and justify the losses due to the temperature difference with the surrounding air. Write a reasoned proposal with two measures to reduce electricity consumption when cooking.

Evaluation criteria:

• Rigor in taking measurements, expressing uncertainties, and clarity of the graph.

• Conceptual correctness in the calculation and distinction between electrical work and transferred heat.

• Physical justification of the efficiency obtained and feasibility of the energy-saving proposals.