Question

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

Answer: The specific heat of granite is approximately 791.89 J/(kg·°C).

Step-by-step solution

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. \(Q = 2.93\,\text{kJ} \times 1000 = 2930\,\text{J}\)

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.

Rearrange the equation to solve for specific heat (c)

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

Plug in the given values and solve for c

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

State the final answer

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