/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 51 The specific heat of iron metal ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The specific heat of iron metal is \(0.450 \mathrm{~J} / \mathrm{g}-\mathrm{K}\). How many J of heat are necessary to raise the temperature of a 1.05-kg block of iron from \(25.0{ }^{\circ} \mathrm{C}\) to \(88.5^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
The heat necessary to raise the temperature of a 1.05-kg block of iron from \(25.0{ }^{\circ} \mathrm{C}\) to \(88.5{ }^{\circ} \mathrm{C}\) can be calculated using the formula \(q = mc\Delta T\). Converting the mass to grams (1050 g) and calculating the change in temperature (63.5 K), we find that the heat absorbed is \(q = (1050 \mathrm{g})(0.450 \mathrm{J/g-K})(63.5 \mathrm{K}) = 29925 \mathrm{J}\). So, 29,925 J of heat are necessary.

Step by step solution

01

Convert the mass from kg to g

In order to convert the mass of the iron block from kg to g, we need to multiply by 1000 since there are 1000 g in 1 kg: mass (g) = 1.05 kg * 1000 (g/kg) = 1050 g
02

Calculate the change in temperature

ΔT is the difference between the final temperature (88.5°C) and the initial temperature (25.0°C). To calculate the change in temperature, subtract the initial temperature from the final temperature: ΔT = 88.5°C - 25.0°C = 63.5 K (Note: The change in temperature is expressed in Kelvin (K) since the specific heat capacity is given in J/g-K. However, the change in temperature is the same if expressed in Celsius degrees (°C). So, we can leave ΔT in K.)
03

Calculate the heat absorbed using the formula

Now that we have the mass in g and the change in temperature in K, we can use the formula to find the heat absorbed: q = mcΔT q = (1050 g)(0.450 J/g-K)(63.5 K) q = 29925 J So, 29,925 J of heat are necessary to raise the temperature of the 1.05-kg block of iron from 25.0°C to 88.5°C.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heat Transfer Calculation
Heat transfer is the process of energy exchange due to a temperature difference. In calculations like this, understanding how to determine the amount of heat exchanged is crucial. The specific heat capacity is the key property that allows us to relate the mass, temperature change, and energy exchanged.

To perform a heat transfer calculation, use the equation:
  • \( q = mc\Delta T \)
  • \( q \) is the heat absorbed or released (in joules).
  • \( m \) is the mass (in grams, once converted).
  • \( c \) is the specific heat capacity (in J/g-K).
  • \( \Delta T \) is the change in temperature (in K or °C).
In our exercise, we applied this formula to find out how much energy is needed to raise the temperature of iron. By plugging in the mass, specific heat, and temperature change, we calculated the heat required. Thus, solving \( q = (1050 \text{ g})(0.450 \text{ J/g-K})(63.5 \text{ K}) = 29925 \text{ J} \).

Understanding heat transfer calculations is helpful in many scientific applications and everyday situations, such as predicting energy use in heating systems.
Temperature Change
Temperature change, symbolized as \( \Delta T \), indicates the difference in the initial and final states of an object's temperature. In physics and chemistry, it represents the driving force behind heat transfer. Calculating temperature change is simple yet essential.

Here's how you can figure it out:
  • Determine the final temperature (\( T_f \)).
  • Identify the initial temperature (\( T_i \)).
  • Subtract the initial from the final temperature: \( \Delta T = T_f - T_i \).
In the context of our problem, the initial temperature is 25.0°C, while the final is 88.5°C, leading to:
\( \Delta T = 88.5 - 25.0 = 63.5 K \) (Kelvin and Celsius changes are equivalent).

The calculated temperature change helps us understand how much energy iron requires to achieve this increase. Even though it seems straightforward, getting this step right is crucial when using formulas involving heat.
Mass Conversion
Converting mass between different units is a common task when performing calculations in physics and chemistry, especially when using formulae in science that expect specific units. In this exercise, converting the iron block's mass from kilograms to grams was needed, as specific heat capacity was provided in J/g-K.

To convert mass from kilograms to grams:
  • Note that 1 kilogram equals 1000 grams.
  • Multiply the number of kilograms by 1000 to convert to grams.
For our iron block:
Mass in kilograms = 1.05 kg
Convert to grams = 1.05 kg * 1000 g/kg = 1050 g.

Converting mass accurately is significant because incorrect units can lead to errors in calculations like heat transfer and others. Consistent unit usage ensures correct results and understanding of material properties.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) What is meant by the term standard conditions, with reference to enthalpy changes? (b) What is meant by the term enthalpy of formation? (c) What is meant by the term standard enthalpy of formation?

A \(1.800-g\) sample of phenol \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\right)\) was burned in a bomb calorimeter whose total heat capacity is \(11.66 \mathrm{~kJ} /{ }^{\circ} \mathrm{C}\). The temperature of the calorimeter plus contents increased from \(21.36^{\circ} \mathrm{C}\) to \(26.37^{\circ} \mathrm{C}\). (a) Write a balanced chemical equation for the bomb calorimeter reaction. (b) What is the heat of combustion per gram of phenol? Per mole of phenol?

From the following data for three prospective fuels, calculate which could provide the most energy per unit volume:$$ \begin{array}{lcc} & \begin{array}{c} \text { Density } \\ \text { at } 20^{\circ} \mathrm{C} \\ \left(\mathrm{g} / \mathrm{cm}^{3}\right) \end{array} & \begin{array}{c} \text { Molar Enthalpy } \\ \text { of Combustion } \\ \text { Fuel } \end{array} & \mathrm{kJ} / \mathrm{mol} \\ \hline \text { Nitroethane, } \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NO}_{2}(l) & 1.052 & -1368 \\ \text { Ethanol, } \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l) & 0.789 & -1367 \\ \text { Methylhydrazine, } \mathrm{CH}_{6} \mathrm{~N}_{2}(l) & 0.874 & -1305 \end{array} $$

Suppose an Olympic diver who weighs \(52.0 \mathrm{~kg}\) executes a straight dive from a \(10-\mathrm{m}\) platform. At the apex of the dive, the diver is \(10.8 \mathrm{~m}\) above the surface of the water. (a) What is the potential energy of the diver at the apex of the dive, relative to the surface of the water? (b) Assuming that all the potential energy of the diver is converted into kinetic energy at the surface of the water, at what speed in \(\mathrm{m} / \mathrm{s}\) will the diver enter the water? (c) Does the diver do work on entering the water? Explain.

(a) Why is the change in enthalpy usually easier to measure than the change in internal energy? (b) For a given process at constant pressure, \(\Delta H\) is negative. Is the process endothermic or exothermic?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.