Question

0.075 kgof water at 10ºC is heated by supplying 25200 J of heat energy. Calculate the final temperature of water.

Answer 1

Answer:

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Explanation:

$to \: find = \\ final \: temperature \: of \: water \\ \\ so \: let \: final \: temperature \: of \: water \: be \: T°C \\ so \: here \\ mass \: of \: water \: (m) = 0.075 \: kg= 75.g \\ heat \: energy \: (Q) = 25200.J \\initial \: temperature \: (t) = 10 °C \\ specific \: heat \: capacity \: of \: water \: (c) = 4.2 \: J/g°C\: \: \\ \\ so \: as \: here \: final \: temperature \: is \: T°C \\ ∆t = (T - 10) \\ so \: we \: know \: that \\ Q = m.c.∆t \\ 25200 = 4.2 \times 75 \times (T-10) \\ 25200 = 315(T-10) \\ T-10 = \frac{25200}{315} \\ T-10 = 80 \\ \\ T = 10 + 80 \\ T = 90 \: °C \\ \\ hence \: final \: temperature \: of \: water \: was \: 90 °C$