Identify the best practices when storing and using drying agents in the lab.
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Ryan calculates that he requires 4 ice cubes for 250mL of hot tea to reach the optimal drinking temperature by using the specific heat capacity of tea.
A substance's potential to hold heat is indicated by its specific heat capacity. This substance size reflects the amount of heat required to raise a certain volume of a material's temperature by one Kelvin. Specific heat capacity is a distinguishing feature of every substance and is useful for material identification.
Given:
Final temperature of system is 57.8℃
dT1 = 79.1 - 57.8 = 21.3℃
dT2 = 0 - (-8.33) = 8.33℃
dT3 = 57.8 - 0 = 57.8℃
Mass of tea, m1 = 250.0mL = 250.0g = 0.250kg
Specific heat capacity of tea, C1 = 4186 J/kg℃ = Specific heat capacity of water
Specific heat capacity of ice, Ci = 2090 J/kg℃
Mass of each ice cubes, m of i = 18.8g
Latent heat of fusion of ice, Lf = 334 kJ/kg
To find:
No. of ice cubes required = ?
Calculations:
Suppose equilibrium temperature is T, then
Heat released by Tea = Heat gained by Ice
Q1 = Q2
m1x C1x dT1 = mi x Ci x dT2 + mi x Lf + mi x Cw x dT3
0.250x 4186 x 21.3 = mi x (2090 x 8.33 + 3.34 x + 4186 x 57.8)
mi = 0.250 x 4186 x 21.3 / (2090 x 8.33 + 3.34 x + 4186 x 57.8)
mi = 0.0761kg = 76.1g
Mass of ice required = 76.1g
Number of ice cube required will be:
n = Total mass of ice/mass of each ice cube
n = 76.1/18.8
n = 4.04 ice cubes = 4.0 ice cubes
Result:
Ryan requires 4 ice cubes to bring 250mL of hot tea to the optimal drinking temperature.
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