An immersion heater of power J= 500 W is used to heat water in a bowl. After 2 minutes, the temperature increases from T1= 85°C to T2 = 90°C. The heater is then switched off for an additional minute, and the temperature drops by Estimate the mass of the water in the bowl. The thermal capacity of water c = 4.2 x 103 J K-1 Kg-1

0.051kg

Explanation:

Using the law of calorimetry which states that the amount of heat lost by a hot substance is equal to the amount of heat gained by the cold substance.

Applying the law to this question,

Amount of heat lost by the heater = heat gained by the water

Heat lost by Immersion heater Q1 = Power×time

Given power = 500W time = 2minutes = 120seconds

Heat loss by immersion heater = 500×120 = 60,000Joules

Heat gained by water = mc∆t where;

m is the mass of the water in the bowl = ?

c is the specific heat capacity = 4.2x10³JK-1 Kg-1

∆t = change in temperature = 90°C -85°C = 5°C

= 273+5 = 278K

Substituting in the formula we have;

Q2= 4.2×10³×278×m = 1,167,600m

According to the law if calorimetry,

Q2=Q1

1,167,600m = 60000

m = 60000/1,167,600

m = 0.051kg

Therefore the mass of water in the bowl is 0.051kg

d eras and periods

the components of a vector are the parts of the vectors that are to each other.

b perpendicular