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Chapter 14: Problem 83

A \(0.500-\mathrm{kg}\) slab of granite is heated so that its temperature increases by \(7.40^{\circ} \mathrm{C} .\) The amount of heat supplied to the granite is \(2.93 \mathrm{kJ} .\) Based on this information, what is the specific heat of granite?

Short Answer

Expert verified
Answer: The specific heat of granite is approximately 791.89 J/(kg路掳C).

Step by step solution

01

Analyze the given information

We have been given the following information: mass of granite, \(m = 0.500\,\text{kg}\) change in temperature, \(\Delta T = 7.40\,{}^{\circ}\,\text{C}\) amount of heat supplied, \(Q = 2.93\,\text{kJ}\) (However, we will need to convert it into joules) Before moving forward, let's convert the heat supplied in kJ to joules (J). To do this, we need to multiply the given amount of heat by 1000. \(Q = 2.93\,\text{kJ} \times 1000 = 2930\,\text{J}\)
02

Use the formula for the heat equation

The formula for the heat equation is: \(Q = mc\Delta T\) where: \(Q\) - amount of heat supplied \(m\) - mass of the substance \(c\) - specific heat of the substance \(\Delta T\) - change in temperature Our goal is to determine \(c\) given the aforementioned information.
03

Rearrange the equation to solve for specific heat (c)

We need to solve for \(c\). To do this, we need to rearrange the formula: \(c = \dfrac{Q}{m\Delta T}\)
04

Plug in the given values and solve for c

Now we can plug in the given values into the rearranged equation: \(c = \dfrac{2930\,\text{J}}{0.500\,\text{kg} \times 7.40\,{}^{\circ}\,\text{C}}\) \(c = \dfrac{2930}{(0.500)(7.40)}\) \(c = \dfrac{2930}{3.7}\) \(c = 791.89\,\dfrac{\text{J}}{\text{kg}\,{}^{\circ}\,\text{C}}\)
05

State the final answer

The specific heat of granite is approximately \(791.89\,\dfrac{\text{J}}{\text{kg}\,{}^{\circ}\,\text{C}}\).

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Most popular questions from this chapter

One end of a cylindrical iron rod of length \(1.00 \mathrm{m}\) and of radius \(1.30 \mathrm{cm}\) is placed in the blacksmith's fire and reaches a temperature of \(327^{\circ} \mathrm{C} .\) If the other end of the rod is being held in your hand \(\left(37^{\circ} \mathrm{C}\right),\) what is the rate of heat flow along the rod? The thermal conductivity of yal iron varies with temperature, but an average between the two temperatures is $67.5 \mathrm{W} /(\mathrm{m} \cdot \mathrm{K})$. (tutorial: conduction)

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