QuestionJune 28, 2025

A sample of ethyl alcohol and a sample of aluminum have the same mass. Which sample absorbs more energy as it is heated from 25^circ C to 56^circ C c_(ethanol)=0.58kcal/kghat (A)^circ C and c_(aluminum)=0.22kcal/kghat (A)^circ C aluminum ethyl alcohol

A sample of ethyl alcohol and a sample of aluminum have the same mass. Which sample absorbs more energy as it is heated from 25^circ C to 56^circ C c_(ethanol)=0.58kcal/kghat (A)^circ C and c_(aluminum)=0.22kcal/kghat (A)^circ C aluminum ethyl alcohol
A sample of ethyl alcohol and a sample of aluminum have the same mass. Which sample absorbs more energy as it is heated from 25^circ C to 56^circ C c_(ethanol)=0.58kcal/kghat (A)^circ C and c_(aluminum)=0.22kcal/kghat (A)^circ C aluminum ethyl alcohol

Solution
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Answer

Ethyl alcohol absorbs more energy. Explanation 1. Identify the formula for heat absorption The energy absorbed is calculated using **Q = mc\Delta T**, where m is mass, c is specific heat capacity, and \Delta T is the change in temperature. 2. Calculate temperature change \Delta T = 56^{\circ}C - 25^{\circ}C = 31^{\circ}C 3. Compare energy absorption for both substances Since mass m is the same for both, compare c \times \Delta T: - Ethyl alcohol: 0.58 \, \text{kcal/kg}^{\circ}C \times 31^{\circ}C = 17.98 \, \text{kcal/kg} - Aluminum: 0.22 \, \text{kcal/kg}^{\circ}C \times 31^{\circ}C = 6.82 \, \text{kcal/kg}

Explanation

1. Identify the formula for heat absorption<br /> The energy absorbed is calculated using **$Q = mc\Delta T$**, where $m$ is mass, $c$ is specific heat capacity, and $\Delta T$ is the change in temperature.<br /><br />2. Calculate temperature change<br /> $\Delta T = 56^{\circ}C - 25^{\circ}C = 31^{\circ}C$<br /><br />3. Compare energy absorption for both substances<br /> Since mass $m$ is the same for both, compare $c \times \Delta T$:<br />- Ethyl alcohol: $0.58 \, \text{kcal/kg}^{\circ}C \times 31^{\circ}C = 17.98 \, \text{kcal/kg}$<br />- Aluminum: $0.22 \, \text{kcal/kg}^{\circ}C \times 31^{\circ}C = 6.82 \, \text{kcal/kg}$
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