If 150 kcal of heat raises the temperature of 2.0 kg of a material by 400 F°, what is the specific heat capacity of the material?

Question

Quizplus · Accepted Answer

The specific heat capacity (c) can be calculated using the formula $q = mc\Delta T$, where $q$ is the heat added, $m$ is the mass, and $\Delta T$ is the temperature change. First, convert the temperature change from Fahrenheit to Celsius, as specific heat capacity is often given in terms of Celsius. The conversion is $\Delta T(°C) = \Delta T(°F) \times \frac{5}{9}$. So, $400°F = 400 \times \frac{5}{9} = 222.22°C$. Then, rearrange the formula to solve for $c$: $c = \frac{q}{m\Delta T}$. Plugging in the values gives $c = \frac{150}{2 \times 222.22} = 0.3375 \mathrm { kcal } / \mathrm { kg } \cdot \mathrm { C } ^ { \circ }$, which is closest to option B, $0.34 \mathrm { kcal } / \mathrm { kg } \cdot \mathrm { C } ^ { \circ }$.

If 150 Kcal of Heat Raises the Temperature of 2 0.75kcal/kg⋅C∘0.75 \mathrm { kcal } / \mathrm { kg } \cdot \mathrm { C } ^ { \circ }0.75kcal/kg⋅C∘

If 150 Kcal of Heat Raises the Temperature of 2 $0.75 \mathrm { kcal } / \mathrm { kg } \cdot \mathrm { C } ^ { \circ }$