A 50g vhuck of metal (c=0.506 j/g degrees celsius) is heated to 60 degrees celsius. It is then dropped into a calorimeter (c=1.40 j/g derees celsius, m - 5.0g) that contains 100g of liquid. The temperature of the liquid raises from 15 degrees celsius to 20 degrees celsius. What is the celsius of the liquid?

To find the final temperature of the liquid, we need to use the heat transfer equation:

q1 = q2

The heat transferred from the metal to the liquid (q1) is equal to the heat absorbed by the liquid (q2).

To calculate q1, we need to use the specific heat formula:

q1 = m1 * c1 * ΔT1

where:
m1 = mass of the metal (50g)
c1 = specific heat of the metal (0.506 J/g°C)
ΔT1 = change in temperature of the metal (60°C - initial temperature of the metal)

To calculate q2, we need to use the specific heat formula:

q2 = m2 * c2 * ΔT2

where:
m2 = mass of the liquid (100g)
c2 = specific heat of the liquid in the calorimeter (1.40 J/g°C)
ΔT2 = change in temperature of the liquid (final temperature of liquid - initial temperature of the liquid)

Since q1 = q2, we can set up the equation:

m1 * c1 * ΔT1 = m2 * c2 * ΔT2

Now, let's substitute the given values:

(50g) * (0.506 J/g°C) * (60°C - initial temperature of the metal) = (100g) * (1.40 J/g°C) * (final temperature of liquid - 15°C)

Simplifying the equation:

25.3 * (60 - initial temperature of the metal) = 140 * (final temperature of liquid - 15)

Now, we can solve for the final temperature of the liquid by rearranging the equation:

final temperature of the liquid = (25.3 * (60 - initial temperature of the metal)) / 140 + 15

Given that the initial temperature of the metal and the final temperature of the liquid are known, we can substitute the values into the equation to find the final temperature of the liquid.