A 0.5 kg object at 16°C is put into a beaker containing 2 kg of water at 25°C. By assuming that no heat is lost to the surrounding, calculate the final temperature of the water. (Specific heat capacity of object = 840J/kg °C, specific heat capacity of water = 4200J/kg°C)

HELP ME! :(

.5 (840)(T-16) = 2(4200)(25-T)

To solve this problem, we can use the concept of heat transfer. The heat gained by the water will be equal to the heat lost by the object:

Heat gained by water = Heat lost by object

The formula to calculate heat is:

Q = mcΔT

Where:
Q is the heat (in joules)
m is the mass (in kilograms)
c is the specific heat capacity (in J/kg°C)
ΔT is the change in temperature (in °C)

Let's calculate the heat gained by the water:

Qwater = mwater * cwater * ΔTwater

Given:
mwater = 2 kg
cwater = 4200 J/kg°C
ΔTwater = Final temperature - Initial temperature = Final temperature - 25°C

Now, let's calculate the heat lost by the object:

Qobject = mobject * cobject * ΔTobject

Given:
mobject = 0.5 kg
cobject = 840 J/kg°C
ΔTobject = Initial temperature - Final temperature = 16°C - Final temperature

Since the heat gained by the water is equal to the heat lost by the object, we can set up the equation:

Qwater = Qobject

Applying the formulas:

mwater * cwater * ΔTwater = mobject * cobject * ΔTobject

Rearranging the equation to solve for the final temperature of the water:

Final temperature = (mwater * cwater * ΔTwater / (mwater * cwater + mobject * cobject)) + Initial temperature

Now, substitute the given values and solve for the final temperature:

Final temperature = (2 kg * 4200 J/kg°C * (Final temperature - 25°C) / ((2 kg * 4200 J/kg°C) + (0.5 kg * 840 J/kg°C))) + 16°C

Simplifying the equation:

Final temperature = (8400 (Final temperature - 25) / (8400 + 420)) + 16

Expanding and simplifying further:

Final temperature = (8400 Final temperature - 210,000) / 8820 + 16

By solving this equation, you can find the final temperature of the water.