Asked by vkny
A small electric immersion heater is used to heat 80 g of water for a cup of instant coffee. The heater is labeled “130 watts” (it converts electrical energy to thermal energy at this rate). Calculate the time required to bring all this water from 24°C to 100°C, ignoring any heat losses. (The specific heat of water is 4186 J/kg·K.)
Answers
Answered by
oobleck
just take care of the units.
80g*(100-24)°C * 1kg/1000g * 4186J/kg-°C * 1W/(1J/s) / 130W
= (80*76*4186)/(1000*130) = 195.776 s
seems like a long time, but 130W isn't much heat.
80g*(100-24)°C * 1kg/1000g * 4186J/kg-°C * 1W/(1J/s) / 130W
= (80*76*4186)/(1000*130) = 195.776 s
seems like a long time, but 130W isn't much heat.
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