Question

70.0 g of water are cooled by removing 900 J of heat. The final temperature is 4°C. What was the initial temperature?A. 5°C. B. 7°C.C. 3°C.D. 6°C.

Answer 1

The initial temperature of water is 7 degrees Celcius

Option B

Step-by-step explanation

Given that;

The mass of water is 70.0 grams

The heat removed is -900J

The final temperature of the water is 4 degrees Celcius

The specific heat capacity of water is 4.182 J/g degrees celcius

To find the initial temperature, follow the steps below

Step 1; Write the formula for calculating heat energy

`\text{ q }=\text{ mc\lparen}\theta_2\text{ - }\theta_1)`

Where

q is the heat

m is the mass of the water

c is the specific heat capacity

theta 1 is the initial temperature

theta 2 is the final temperature

Step 2; Substitute the above given data into the formula in step 1

`\begin{gathered} \text{ -900 }=\text{ 70 }*\text{ 4.182 \lparen4 - }\theta_1) \\ -\text{ 900 }=\text{ 292.81 \lparen4 - }\theta_1) \\ \text{ -900 }=\text{ 1,171.24 - 292.81}\theta_1 \\ \text{ Subtract 1,171.24 from both sides} \\ \text{ -900 - 1,171.24 }=\text{ 1,171.24 - 1,171.24 - 292.81}\theta_1 \\ \text{ -2,071.24 }=\text{ -292.8}\theta_1 \\ \text{ Divide both sides by -292.8} \\ \text{ }(-2,071.24)/(-292.8)=\frac{-292.8\text{ }\theta_1}{-292.8} \\ \text{ }\theta_1\text{ }=\text{ 7.08}\degree C \\ \theta_(_1)\approx\text{ 7}\degree C \end{gathered}`

Hence, the initial temperature of water is 7 degrees Celcius