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Two bars of identical mass are at \(25^{\circ} {C}\) . One is made from glass and the other from another substance. The specific heat capacity of glass is 840 \({J} /({kg} \cdot {C}^{\circ})\) . When identical amounts of heat are supplied to each, the glass bar reaches a temperature of \(88^{\circ} {C}\) , while the other bar reaches 250.0 'C. What is the specific heat capacity of the other substance?

Short Answer

Expert verified
The specific heat capacity of the other substance is 235.2 J/(kg·°C).

Step by step solution

01

Understand the Concept of Heat Transfer

The problem involves the concept of specific heat capacity, which is the amount of heat required to change the temperature of a substance by 1 degree Celsius per unit mass. When identical amounts of heat are supplied to different substances, their temperature changes depending on their specific heat capacities.
02

Apply the Formula for Heat Transfer

The formula for heat transfer is given by: \[ Q = mc\Delta T \]where \( Q \) is the heat energy supplied, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature. For both bars, the heat supplied \( Q \) is the same.
03

Calculate Change in Temperature for Glass

For the glass bar, the initial temperature \( T_i = 25^{\circ} C \) and the final temperature \( T_f = 88^{\circ} C \). The change in temperature is:\[ \Delta T_{glass} = T_f - T_i = 88 - 25 = 63^{\circ} C \]
04

Calculate Change in Temperature for Other Substance

For the other substance, the initial temperature \( T_i = 25^{\circ} C \) and the final temperature \( T_f = 250^{\circ} C \). The change in temperature is:\[ \Delta T_{other} = T_f - T_i = 250 - 25 = 225^{\circ} C \]
05

Set Up the Heat Balance Equation

Since the amount of heat \( Q \) is the same for both bars:\[ m c_{glass} \Delta T_{glass} = m c_{other} \Delta T_{other} \]Notice that the mass \( m \) cancels out as it is the same for both bars.
06

Solve for Specific Heat Capacity of the Other Substance

We are given \( c_{glass} = 840 \ J/(kg \cdot C^{\circ}) \).Using the heat balance equation:\[ c_{glass} \Delta T_{glass} = c_{other} \Delta T_{other} \]\[ 840 \times 63 = c_{other} \times 225 \]Solve for \( c_{other} \):\[ c_{other} = \frac{840 \times 63}{225} = 235.2 \ J/(kg \cdot C^{\circ}) \]

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Key Concepts

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

Heat Transfer
Heat transfer is the process of energy moving from one place or material to another. This movement occurs due to a difference in temperature between the two areas or substances involved. Heat will always transfer or flow from a higher temperature to a lower temperature, working to balance the thermal energy between them.
In our exercise, identical amounts of heat are applied to two bars of equal mass, made of different materials. This scenario helps us observe how different substances react to the same amount of heat due to their intrinsic properties. Each bar receives the same amount of thermal energy, but due to different specific heat capacities, they experience varying changes in temperature. Understanding this concept is key to explaining why substances behave differently when subjected to heat.
Temperature Change
A temperature change occurs when heat energy is added to or removed from a substance, causing its temperature to increase or decrease, respectively. The amount of temperature change depends on how much heat is added or removed and the substance's specific heat capacity.
In the exercise, both the glass bar and the other substance initially have a temperature of 25°C. After the heat is applied, the glass bar reaches 88°C and the other bar gets as hot as 250°C. By calculating the difference between the final and initial temperatures for each material, we learn how much their temperatures increased.
For the glass, the temperature change or \(\Delta T\) is 63°C \((88 - 25)\). For the other substance, \(\Delta T\) is 225°C \((250 - 25)\). Recognizing these temperature shifts helps us determine how each material's specific heat capacity influences their response to heat.
Heat Energy Equation
The heat energy equation is a central tool in thermodynamics, used to calculate the amount of heat transferred to or from an object. The equation is formulated as: \[ Q = mc\Delta T \] where
  • \( Q \) is the heat energy (in joules),
  • \( m \) is the mass (in kilograms),
  • \( c \) is the specific heat capacity (in joules per kilogram per degree Celsius),
  • \( \Delta T \) is the change in temperature (in degrees Celsius).
In the given problem, this equation helps us express the relationship between the heat supplied and the resulting change in temperature for different materials.
Since both bars received the same heat \( Q \), we can set their heat energy equations equal to each other to derive the specific heat capacity of the unknown substance. By isolating this variable, we discover that the other substance has a specific heat capacity of 235.2 \( J/(kg \cdot C^{\circ}) \). This demonstrates how the heat energy equation allows us to calculate unknown properties when enough variables are provided.

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Most popular questions from this chapter

The vapor pressure of water at \(10^{\circ} {C}\) is 1300 \({Pa}\) . (a) What percentage of atmospheric pressure is this? Take atmospheric pressure to be \(1.013 \times 10^{5} {Pa}\) . (b) What percentage of the total air pressure at \(10^{\circ} {C}\) is due to water vapor when the relative humidity is 100\(\% ?\) (c) The vapor pressure of water at \(35^{\circ} {C}\) is 5500 Pa. What is the relative humidity at this temperature if the partial pressure of water in the air has not changed from what it was at \(10^{\circ} {C}\) when the relative humidity was 100\(\% ?\)

A copper–constantan thermocouple generates a voltage of \(4.75 \times 10^{-3}\) volts when the temperature of the hot junction is \(110.0^{\circ} \mathrm{C}\) and the reference junction is kept at a temperature of 0.0 C. If the voltage is proportional to the difference in temperature between the junctions, what is the temperature of the hot junction when the voltage is \(1.90 \times 10^{-3}\) volts?

ssm mmh A 42 -kg block of ice at \(0^{\circ} {C}\) is sliding on a horizontal surface. The initial speed of the ice is 7.3 m/s and the final speed is 3.5 m/s. Assume that the part of the block that melts has a very small mass and that all the heat generated by kinetic friction goes into the block of ice. Determine the mass of ice that melts into water at \(0^{\circ} {C}\)

Blood can carry excess energy from the interior to the surface of the body, where the energy is dispersed in a number of ways. While a person is exercising, 0.6 kg of blood flows to the body’s surface and releases 2000 J of energy. The blood arriving at the surface has the temperature of the body’s interior, \(37.0^{\circ} {C}\) . Assuming that blood has the same specific heat capacity as water, determine the temperature of the blood that leaves the surface and returns to the interior.

What’s your normal body temperature? It may not be 98.6 \(^{\circ} \mathrm{F}\), the often-quoted average that was determined in the nineteenth century. A more recent study has reported an average temperature of 98.2 \(^{\circ} \mathrm{F}\). What is the difference between these averages, expressed in Celsius degrees?

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