Question

Gold cylinder has a mass of 75 g and a specific heat of 0.129J/G degrees Celsius it is heated to 65°C and then put in 500 g of water who’s temperature is 90°C what is the final temperature of the gold water system

Answer 1

89.88° C

Step-by-step explanation:

We are given;

• Mass of gold cylinder as 75 g
• specific heat of gold is 0.129 J/g°C
• Initial temperature of gold cylinder is 65°C
• Mass of water is 500 g
• Initial temperature of water is 90 °C

We are required to calculate the final temperature;

• We know that Quantity of heat is given by the product of mass, specific heat capacity and change in temperature.

• That is, Q = m × c × ΔT

Step 1: Calculate the quantity of heat absorbed by the Gold cylinder

Assuming the final temperature is X° C

Then; ΔT = (X-65)°C

Therefore;

Q = 75 g × 0.129 J/g°C × (X-65)°C

= 9.675X - 628.875 Joules

Step 2: Calculate the quantity of heat released by water

Taking the final temperature as X° C

Change in temperature, ΔT = (90 - X)° C

Specific heat capacity of water is 4.184 J/g°C

Therefore;

Q = 500 g × 4.184 J/g°C × (90 - X)° C

= 188,280 -2092X joules

Step 3: Calculate the final temperature, X°C

we know that the heat gained by gold cylinder is equal to the heat released by water.

9.675X - 628.875 Joules = 188,280 -2092X joules

2101.675 X = 188908.875

X = 89.88° C

Thus, the final temperature is 89.88° C