Question

exercices supplémentaires

1. on verse 250 ml deau dans un verre sortant du congélateur. après quelque temps, on constate que la température de leau est passée de 18 °c à 12 °c. quelle quantité de chaleur a été transférée entre leau et le verre ?

Explanation:

Step1: Identify the formula for heat transfer

The formula for heat transfer (when there is no phase change) is \( Q = mc\Delta T \), where \( Q \) is the heat transferred, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature. For water, the density \(
ho = 1\ g/ml \), so for \( 250\ ml \) of water, the mass \( m =
ho V = 1\ g/ml\times250\ ml = 250\ g = 0.25\ kg \). The specific heat capacity of water \( c = 4186\ J/(kg\cdot^{\circ}C) \). The change in temperature \( \Delta T = T_{final}-T_{initial}=12^{\circ}C - 18^{\circ}C=- 6^{\circ}C \) (the negative sign indicates heat is lost by water, gained by the glass, but we can consider the magnitude for the amount transferred).

Step2: Calculate the heat transfer

Substitute the values into the formula: \( Q = mc\Delta T \). Plugging in \( m = 0.25\ kg \), \( c = 4186\ J/(kg\cdot^{\circ}C) \), and \( \Delta T = 6^{\circ}C \) (taking the absolute value for the amount of heat transferred), we get \( Q=0.25\ kg\times4186\ J/(kg\cdot^{\circ}C)\times6^{\circ}C \). First, calculate \( 0.25\times6 = 1.5 \), then \( 1.5\times4186 = 6279\ J \approx 6280\ J \) or \( 6.28\ kJ \). (Note: If we consider the sign, the negative value would indicate heat lost by water, but the question asks for the quantity of heat transferred between water and glass, so the magnitude is what's important here.)

Answer:

The quantity of heat transferred is approximately \( 6280\ J \) (or \( 6.28\ kJ \)) (assuming the specific heat capacity of water is \( 4186\ J/(kg\cdot^{\circ}C) \) and density of water is \( 1\ g/ml \)).