Question

What is the final temperature of water given an initial temperature of 28 ˚C, a mass of 7 g, and heat (q) of 184 J (Specific Heat of water = 4.184 )?

Answer 1

Answer:34.28 Degree celcius

Explanation:The heat given is equal to the change in temperature multiplied by the mass and the specific heat capacity

This gives the equation:Q=mc∆T

Where Q=184,m=7, specific heat capacity=4.184,∆T=X-28,

But ∆T=Final temperature (T'')-Initial temperature (T')

Therefore the equation becomes

184=7×4.184(X-28)

Solving the remaining linear equation gives you X as 34.28°C