Water of mass 120g at 50oC is added to 20og of water at 10oC and the mixture is well stirred. Calculate the temperature of the mixture. [Neglect heat losses to the surrounding]
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option C
Explanation:
120 x 4.2(50 - \(\theta\)) = 200 x 4.2(\(\theta\) - 10)
\(\theta = \frac{33600}{1344} = 25^oC\)
120 x 4.2(50 - \(\theta\)) = 200 x 4.2(\(\theta\) - 10)
\(\theta = \frac{33600}{1344} = 25^oC\)