A 100 g glass container contains 250 g of water at 15.0 °C. A 100 g piece of unknown material at 100 °C is added to the water in the container. The final temperature of the mixture is 19.0 °C. What is the specific heat of the unknown material in J/kg °C? (specific heat of water = 4,186 J/kg °C, specific heat of glass = 837.2 J/kg °C)

Question

Quizplus · Accepted Answer

First, we need to find the amount of heat lost by the hot unknown material as it cools down to the final temperature, and the amount of heat gained by the cold water as it warms up to the final temperature. These two amounts of heat must be equal since no heat is lost or gained by the container itself, as it stays at the same temperature throughout the process.

The formula for heat transfer is: 
q = mcΔT

where 
q = amount of heat transferred 
m = mass of the substance 
c = specific heat of the substance 
ΔT = change in temperature

For the unknown material: 
q1 = (0.1 kg)(c1)(100°C - 19°C) = 8.1c1

For the water: 
q2 = (0.25 kg)(4186 J/kg°C)(19°C - 15°C) = 418.6 J

Since q1 = q2: 
8.1c1 = 418.6

c1 = 51.7 J/kg°C

However, we need to take into account the fact that the container is made of glass, which also gains some heat from the water and unknown material. The amount of heat gained by the glass is:

q3 = (0.1 kg)(837.2 J/kg°C)(19°C - 15°C) = 334.88 J

So the net heat gained by the water and the glass combined is:

q4 = q2 + q3 = 418.6 J + 334.88 J = 753.48 J

The specific heat of the unknown material can then be calculated as:

c1 = q4/(0.35 kg)(100°C - 19°C) = 558 J/kg°C

Therefore, the answer is D) 558.

A 100 G Glass Container Contains 250 G of Water