Question

3 kg of water (c = 4,186 J /kg°C) is at an initial temperature of 10°C. If 7,700 J of heat is applied, what will be its final temperature? Show all of your work and use the correct units for full credit.

Answer 1

`\bold{\huge{\underline{ Solution }}}`

Given :-

• Here , The initial temperature of 3kg of water is 10° C
• The heat supplied to 3kg water is 7,700 J
• The specific heat of water is 4186 J/kg ° C

To Find :-

• We have to find the final temperature of water .

Let's Begin :-

Here, we have

• Mass = 3 kg
• Specific heat capacity = 4186 J/kg°C
• Initial temperature = 10° C
• Heat applied = 7,700J

We know that,

• Specific heat is the heat that is required to increase the temperature of 1 kg or unit mass by 1° C or unit ° C

That is,

`\bold{\red{ C = }}{\bold{\red{\frac{ Q}{m{\delta}T}}}}`

• Here, Change in temperature is ΔT that is, Final temperature - Initial temperature

So,

`\sf{{\delta} T = }{\sf{( Q)/(Cm)}}`

Subsitute the required values,

`\sf{ T2 - 10 = }{\sf{\frac{ 7700}{3{*}4186}}}`

`\sf{ T2 - 10 = }{\sf{( 7700)/(12558)}}`

`\sf{ T2 - 10 = }{\sf{\cancel{( 7700)/(12558)}}}`

`\sf{ T2 - 10 = 0.61 }`

`\sf{ T2 = 0.61 + 10 }`

`\bold{ T2 = 10.61 {\degree} C}`

Hence, The final temperature of 3kg if 7,700 J of heat supplied is 10.61 °C