How much heat is required to increase the temperature of 0.5 kg of water from 10°C to 65°C? (115500 J)
Difficulty: Easy
Solution: Mass of water = m = 0.5 kg
Initial temperature = $T_{1}$ = 10°C = 10 + 273 = 283 K
Final temperature = $T_{2}$ = 65°C = 65 + 273 = 338 K
Change in temperature = $\triangle$ T = $T_{2} - T_{1}$ = 338 – 283 = 55 K
Heat =$ \triangle Q$ = ?
$\triangle Q$ = $mc \triangle T$
$\triangle Q = 0.5\times 2400 \times55$
$\triangle Q$ = 115500J
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