Question

778- j of heat is absorbed by 155 g of water at 34.2 degrees celcius. What is the temperature of the wwater after absorbing the heat

Answer 1

To find the final temperature of the water after absorbing 778 joules of heat, we can use the equation for heat transfer:

Q = m * c * ΔT

where: Q is the heat absorbed (in joules), m is the mass of the water (in grams), c is the specific heat capacity of water (in joules per gram per degree Celsius), and ΔT is the change in temperature (in degrees Celsius).

We are given the following information: Q = 778 J (heat absorbed) m = 155 g (mass of water) c = 4.18 J/(g°C) (specific heat capacity of water at approximately 25°C)

We want to find ΔT (change in temperature). Let's rearrange the equation to solve for ΔT:

ΔT = Q / (m * c)

Now, plug in the given values:

ΔT = 778 J / (155 g * 4.18 J/(g°C))

ΔT ≈ 778 J / (648.9 J/°C)

ΔT ≈ 1.198°C

Now, to find the final temperature of the water, we add the change in temperature (ΔT) to the initial temperature (34.2°C):

Final Temperature = Initial Temperature + ΔT

Final Temperature ≈ 34.2°C + 1.198°C

Final Temperature ≈ 35.398°C

The final temperature of the water after absorbing 778 joules of heat is approximately 35.398°C.