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If 250 grams of water is to be heated from 24.0°C to 100.0°C to make a cup of tea, how much
heat must be added? The specific heat of water is 4.18 J/g∙C

Sagot :

Sarah06109

Answer:

[tex]\boxed {\boxed {\sf 79,420 \ J}}[/tex]

Explanation:

We are given the mass, a change in temperature, and the specific heat of water. We should use the following formula to solve this problem.

[tex]q=mc\Delta T[/tex]

In this formula, m is the mass, c is the specific heat, and ΔT is the change in temperature.

We know there are 250 grams of water and the specific heat of water is 4.18 J/g · °C.

We are given two temperature, so have to find the change in temperature. This is the difference between the initial temperature and final temperature. The water is heated from 24 °C to 100 °C. Therefore, the initial is 24 and the final is 100.

[tex]\bullet \ \Delta T= T_{final} - T_{initial} \\\bullet \ \Delta T=100 \textdegree C - 24 \textdegree C\\\bullet \Delta T= 76 \textdegree C[/tex]

Now we know all three of the variables and we can substitute them into the formula.

[tex]\bullet \ m= 250 \ g\\ \bullet \ c=4.18 \ J/g \textdegree C \\ \bullet \ \Delta T= 76 \textdegree C[/tex]

[tex]q= (250 \ g)( 4.18 \ J/g * \textdegree C)( 76 \textdegree C)[/tex]

Multiply the first two numbers together. The units of grams will cancel.

[tex]q= (1045 \ J/\textdegree C)(76 \textdegree C)[/tex]

Multiply again. This time, the units of degrees Celsius cancel.

[tex]q= 79420 \ J[/tex]

79, 420 Joules of heat must be added.

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