You have 300 g of coffee at 55 degrees celcius. Coffee has the same specific heat as water. How much 10 degrees celcius water do you need to add in order to reduce the coffee's temperature to a more bearable 49 degrees celcius?

Heat gain by water = Heat loss by coffee

mCΔt = mCΔt where C cancel each other

m(49-10) = 300(55-49)

m= 46.154g

To solve this problem, we can use the concept of heat transfer. The heat lost by the coffee as it cools down will be equal to the heat gained by the water as it warms up.

First, let's calculate the heat lost by the coffee:
Q1 = mcΔT

Where:
Q1 represents the heat lost (in joules)
m represents the mass of coffee (in grams)
c represents the specific heat capacity of coffee (which is the same as water, approximately 4.18 J/(g°C))
ΔT represents the change in temperature of the coffee (in °C)

Given:
m = 300 g
c = 4.18 J/(g°C)
ΔT = 55°C - 49°C = 6°C

Plugging in these values into the equation, we have:
Q1 = (300 g)(4.18 J/(g°C))(6°C)
Q1 = 7536 J

Now, let's calculate the heat gained by the water:
Q2 = mcΔT

Where:
Q2 represents the heat gained by the water (in joules)
m represents the mass of water (in grams)
c represents the specific heat capacity of water (approximately 4.18 J/(g°C))
ΔT represents the change in temperature of the water (in °C)

We need to find the mass of water (m) that will absorb this amount of heat (Q1) and decrease its temperature from 10°C to 49°C.

Q2 = Q1

mcΔT = (300 g)(4.18 J/(g°C))(6°C)

Simplifying the equation:
mc = (300 g)(4.18 J/(g°C))(6°C)
mc = 7536 J

Now, let's solve for the mass of water (m):
m = (7536 J) / [(10°C - 49°C)(4.18 J/(g°C))]
m = (7536 J) / (-39°C)(4.18 J/(g°C))
m ≈ -48.46 g

The negative value for mass doesn't make sense in this case. It implies that the water would need to have a negative mass to achieve the desired result, which is not physically possible.

Thus, we cannot add a negative mass of water, and it is not possible to reduce the coffee's temperature from 55°C to 49°C by adding 10°C water.