The water bottle contains 575 grams of water at 80°C. The water eventually cools to 40°C. How many calories of heat could be transferred to the student's back?

–23000 Calories.

NOTE : The negative sign indicates that heat has been loss to the student back.

Explanation:

The following data were obtained from the question:

Mass (M) of water = 575 g

Initial temperature (T1) = 80 °C

Final temperature (T2) = 40 °C

Heat (Q) transferred =.?

Next, we shall determine the change in temperature (ΔT).

This is illustrated below:

Initial temperature (T1) = 80 °C

Final temperature (T2) = 40 °C

Change in temperature (ΔT) =?

Change in temperature (ΔT) = T2 – T1

Change in temperature (ΔT) = 40 – 80

Change in temperature (ΔT) = –40°C

Finally, we shall determine the heat transferred. This can be obtained as follow:

Mass (M) of water = 575 g

Change in temperature (ΔT) = –40°C

Specific heat capacity (C) of water = 1 Cal/g°C

Heat (Q) transferred =.?

Q = MCΔT

Q = 575 × 1 × –40

Q = –23000 Calories

NOTE : The negative sign indicates that heat has been loss to the student back.

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