100-g sample of substance is heated to 100˚C and placed into calorimeter cup containing 150 g of water at 25˚C. The sample raises the temperature of water to 32.1. What is the specific heat of the substance?

The amount of heat transfered to the water is 150*7.1*1.0 = 1065 calories.

Removing that amount of heat from 100 g of the substance caused its temperature to drop 67.9 C. The specific heat can be calculated from that information.

C = Q/(M*deltaT) = 0.157 Cal/gm C
= 0.657 J/g K

To find the specific heat of the substance, we can use the formula:

q = m × c × ΔT

Where:
- q is the heat transferred
- m is the mass of the substance or water
- c is the specific heat capacity
- ΔT is the change in temperature

In this case, we are given:
- The mass of the substance is 100 g
- The initial temperature of the substance is 100 ˚C
- The mass of the water is 150 g
- The initial temperature of the water is 25 ˚C
- The final temperature of the water is 32.1 ˚C

First, we need to calculate the heat transferred to the water.

q_water = m_water × c_water × ΔT_water

Since we want to calculate the specific heat of the substance, we can rewrite the equation as:

q_water = m_substance × c_substance × ΔT_substance

Now, we can rearrange the equation to solve for c_substance:

c_substance = q_water / (m_substance × ΔT_substance)

To find q_water, we can use the equation:

q_water = m_water × c_water × ΔT_water

Where:
- m_water is the mass of water
- c_water is the specific heat capacity of water (which is 4.18 J/g˚C)
- ΔT_water is the change in temperature of the water

ΔT_water = final temperature of the water - initial temperature of the water
ΔT_water = 32.1 ˚C - 25 ˚C
ΔT_water = 7.1 ˚C

m_water = 150 g (given)
c_water = 4.18 J/g˚C (specific heat of water)

Now we can calculate q_water:

q_water = m_water × c_water × ΔT_water
= 150 g × 4.18 J/g˚C × 7.1 ˚C
= 4416.9 J

Now we can plug this value into the equation for c_substance:

c_substance = q_water / (m_substance × ΔT_substance)
= 4416.9 J / (100 g × (100 ˚C - 32.1 ˚C))
= 4416.9 J / (100 g × 67.9 ˚C)
= 6.483 J/g˚C

Therefore, the specific heat of the substance is 6.483 J/g˚C.

To find the specific heat of the substance, we can use the formula:

Q = m * c * ΔT

Where:
Q is the heat gained or lost by the water
m is the mass of the water
c is the specific heat capacity of the substance
ΔT is the change in temperature of the water

In this case, the heat gained by the water can be calculated as:

Q = m_water * c_water * ΔT_water

The mass of the water is given as 150 g, the specific heat capacity of water (c_water) is approximately 4.18 J/g°C, and the change in temperature (ΔT_water) is the final temperature of the water (32.1°C) minus the initial temperature of the water (25°C).

Q = 150 g * 4.18 J/g°C * (32.1°C - 25°C)

Now, we need to determine the heat lost by the substance:

Q = m_substance * c_substance * ΔT_substance

The mass of the substance is given as 100 g, and the initial temperature of the substance is 100°C. The final temperature of the water (32.1°C) is considered the final temperature of the substance (ΔT_substance).

Q = 100 g * c_substance * (32.1°C - 100°C)

Since the heat gained by the water is equal to the heat lost by the substance (Q_water = -Q_substance), we can equate the two equations and solve for c_substance:

150 g * 4.18 J/g°C * (32.1°C - 25°C) = 100 g * c_substance * (32.1°C - 100°C)

To find the specific heat of the substance, we can rearrange the equation:

c_substance = (150 g * 4.18 J/g°C * (32.1°C - 25°C)) / (100 g * (32.1°C - 100°C))

Calculating the equation will give us the specific heat of the substance. The specific heat capacity is a property that varies depending on the substance, so it's important to use the specific values given in the problem.