Question

A hot piece of copper was dropped into 155 g of water at 23.6 °C and 2,100 J of energy was transferred to the water. What is the final temperature of the water?​

Answer 1

The final temperature of the water can be calculated using the formula for heat transfer between copper and water.

Step-by-step explanation:

To calculate the final temperature of the water, we can use the formula:

m1c1(T1 - T) = m2c2(T - T2)

where m1 and m2 are the masses of the copper and water respectively, c1 and c2 are the specific heat capacities of copper and water respectively, T1 is the initial temperature of the copper, T2 is the initial temperature of the water, and T is the final temperature.

Plugging in the given values:

m1 = 248 g

c1 = 0.39 J/g°C (specific heat capacity of copper)

T1 = 314°C

m2 = 390 g

c2 = 4.18 J/g°C (specific heat capacity of water)

T2 = 22.6°C

We can rearrange the formula to solve for T:

T = (m1c1T1 + m2c2T2) / (m1c1 + m2c2)

Plugging in the values:

T = (248 g * 0.39 J/g°C * 314°C + 390 g * 4.18 J/g°C * 22.6°C) / (248 g * 0.39 J/g°C + 390 g * 4.18 J/g°C)

Calculating this, we find that the final temperature of the water is approximately 29.4°C.

Answer 2

26.8 °C

Step-by-step explanation:

Step 1: Given and required data

• Mass of water (m): 155 g

• Energy transferred to the water (Q): 2100 J

• Initial temperature of the water (T₂): 23.6 °C

• Final temperature of the water (T₁): ?

• Specific heat of water (c): 4.184 J/g.°C

Step 2: Calculate the final temperature of the water

We will use the following expression.

Q = c × m × (T₂ - T₁)

T₂ = 2100 J/(4.184 J/g.°C) × 155 g + 23.6 °C = 26.8 °C