Need help with physics question?

moody_134 · #1 (**permalink**) 03-30-2008, 10:58 PM

A small electric immersion heater is used to heat 54 g of water for a cup of instant coffee. The heater is labeled “170 watts” (it converts electrical energy to thermal energy at this rate). Calculate the time required to bring all this water from 18°C to 100°C, ignoring any heat losses. (The specific heat of water is 4186 J/kg·K.)

Blue Sky · #2 (**permalink**) 03-30-2008, 10:58 PM

From 18C to 100C = 291.15K to 373.15K
1 Watt = 1 Joule/s, so 1 W*s = 1 Joule

4186J/kg K (373.15K - 291.15K) (1 kg/1000g)(54g)=
4186W*s/kgK(82K)(0.54kg)

4186Ws/kgK(82K)(0.54kg)/170W = 109 seconds

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