Question

Interactive Solution 12.51 deals with one approach to solving problems such as this. A 0.268-kg coffee mug is made from a material that has a specific heat capacity of 974 J/kg C° and contains 0.298 kg of water. The cup and water are at 27.1 °C. To make a cup of coffee, a small electric heater is immersed in the water and brings it to a boil in two minutes. Assume that the cup and water always have the same temperature and determine the minimum power rating of this heater.

Answer 1

Mass of the coffee mug = m_c = 0.268 kg

Mass of water = m_w = 0.298 kg

Initial temperature of the cup and water = T₁ = 27.1 ^oC

Boiling temperature of water = Final temperature of the cup and water = T₂ = 100 ^oC

Specific heat of the coffee mug = C_c = 974 J/(kg.^oC)

Specific heat of water = C_w = 4186 J/(kg.^oC)

Power rating of the heater = P

Time taken by the heater to raise the temperature = t = 2 min = 120 sec

Pt = m_cC_c(T₂ - T₁) + m_wC_w(T₂ - T₁)

P(120) = (0.268)(974)(100 - 27.1) + (0.298)(4186)(100 - 27.1)

P = 916.39 W

The minimum power rating of the heater = 916.39 W