Question

What is the final temperature after 840 joules is absorbed by 10.0g of water at 25.0 c?

Answer 1

final temperature of water would be 45°C

given, mass of water, m = 10g

initial temperature, T_i = 25°C

final temperature, T_f = ?

Heat , Q = 840 Joule

using formula, Q = mS∆T

⇒840 = 10 × 4.2 J/g/°C × (T_f - 25°C)

⇒20 = T_f - 25°C

⇒T_f = 20 + 25 = 45°C

so the final temperature is 45°C

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Answer 2

$\huge\bold\purple{Answer:-}$

given, mass of water, m = 10g

initial temperature, T_i = 25°C

final temperature, T_f = ?

Heat , Q = 840 Joule

using formula, Q = mS∆T

⇒840 = 10 × 4.2 J/g/°C × (T_f - 25°C)

⇒20 = T_f - 25°C

⇒T_f = 20 + 25 = 45°C

so the final temperature is 45°C