Answer:
–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.
