How much heat is required to warm 122 g of water by 23.0 c?
2 answers:
Answer: 11745.92 j
Explanation: using heat = mc∆
Where k is mass
C is heat capacity of water = 4.186 j/gc
∆ is temperature
H = 122 × 4.186 × 23
= 11745.92 j
<span>122 g * 4,186 (j/g*°c) * 23°c = 11745.916 j </span>
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