Question

Chromium has a specific heat of \(0.450 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\). How much heat is absorbed by \(35.0 \mathrm{~g}\) of chromium if the temperature increases from \(45^{\circ} \mathrm{F}\) to \(88^{\circ} \mathrm{F} ?\)

Short answer

(Specific heat of chromium: 0.450 J/g°C) Answer: Approximately 300.21 J.

Step-by-step solution

Convert given temperatures to Celsius

First, we need to convert both temperatures in Fahrenheit to Celsius. We can do this using the conversion formula: $$T_{C} = \frac{5}{9} \times (T_{F} - 32)$$ \(T_{1\$ =\frac{5}{9} \times(45^{\circ}\mathrm{F}-32)\) \(T_{1C} =\frac{5}{9} \times 13^{\circ}\mathrm{F}\) \(T_{1C} = 7.22^{\circ}\mathrm{C}\) \(T_{2\$ =\frac{5}{9} \times(88^{\circ}\mathrm{F}-32)\) \(T_{2C} =\frac{5}{9} \times 56^{\circ}\mathrm{F}\) \(T_{2C} = 31.11^{\circ}\mathrm{C}\)

Calculate the temperature change ΔT in Celsius

Now calculate the temperature difference (ΔT) in Celsius by subtracting the initial temperature from the final temperature, using the temperatures converted to Celsius: $$\Delta T = T_{2C} - T_{1C}$$ \(\Delta T = 31.11^{\circ}\mathrm{C} - 7.22^{\circ}\mathrm{C}\) \(\Delta T = 23.89^{\circ}\mathrm{C}\)

Calculate the heat absorbed using specific heat, mass, and temperature change

Now, we can find the amount of heat absorbed by the chromium sample using the specific heat formula: $$q = mc\Delta T$$ Where: - q is the heat absorbed. - m is the mass of the chromium (35.0 g). - c is the specific heat of chromium (0.450 J/g°C). - ΔT is the temperature change in Celsius (23.89°C). $$q = 35.0 \mathrm{~g} \times 0.450 \frac{\mathrm J}{\mathrm g \cdot{ }^{\circ}\mathrm{C} } \times 23.89^{\circ}\mathrm{C}$$ $$q = 300.21 \mathrm{J}$$

Conclusion

The amount of heat absorbed by the 35.0g of chromium when its temperature increases from \(45^{\circ} \mathrm F\) to \(88^{\circ} \mathrm F\) is approximately 300.21 J.