QuestionApril 24, 2025

What is q if 28.6 g of water is heated from 22.0^circ C to 78.3^circ C The specific heat of water is 4.184J/gcdot ^circ C 2.60 J 2.63 kJ 6.74 kJ 9.37 kJ 3.94times 10^4kJ

What is q if 28.6 g of water is heated from 22.0^circ C to 78.3^circ C The specific heat of water is 4.184J/gcdot ^circ C 2.60 J 2.63 kJ 6.74 kJ 9.37 kJ 3.94times 10^4kJ
What is q if 28.6 g of water is heated from 22.0^circ C to 78.3^circ C The specific heat of water is 4.184J/gcdot ^circ C 2.60 J 2.63 kJ 6.74 kJ 9.37 kJ 3.94times 10^4kJ

Solution
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Answer

6.74 kJ Explanation 1. Use the heat transfer formula The formula is q = m \cdot c \cdot \Delta T, where m is mass, c is specific heat, and \Delta T is temperature change. 2. Calculate \Delta T \Delta T = 78.3 - 22.0 = 56.3^{\circ}C 3. Substitute values into the formula q = 28.6 \cdot 4.184 \cdot 56.3 = 6740.7 \, J 4. Convert to kJ q = 6740.7 \, J \div 1000 = 6.74 \, kJ

Explanation

1. Use the heat transfer formula<br /> The formula is $q = m \cdot c \cdot \Delta T$, where $m$ is mass, $c$ is specific heat, and $\Delta T$ is temperature change.<br /><br />2. Calculate $\Delta T$<br /> $\Delta T = 78.3 - 22.0 = 56.3^{\circ}C$<br /><br />3. Substitute values into the formula<br /> $q = 28.6 \cdot 4.184 \cdot 56.3 = 6740.7 \, J$<br /><br />4. Convert to kJ<br /> $q = 6740.7 \, J \div 1000 = 6.74 \, kJ$
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