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दिलेले: m = 5 kg, c = 1 kcal/kg °C, सुरुवातीचे तापमान \(T_1\) = 20 °C, अंतिम तापमान \(T_2\) = 100 °C

सूत्र: \(Q = m \times c \times \Delta T\), येथे \(\Delta T = T_2 - T_1\)

उकल: \(\Delta T = 100 - 20 = 80\) °C

\(Q = 5 \times 1 \times 80 = 400\ \text{kcal}\)

उत्तर: पाण्याचे तापमान वाढविण्यासाठी 400 kcal उष्णता लागेल. (ज्यूलमध्ये \(400 \times 4180 = 1.672 \times 10^6\) J)

Doubtly AI · 4 Nov 2026 AI-verified · Nov 2026 Report
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