A 200-L electric water heater uses 2.0 kW. Assuming no heat loss, how many hours would it take to heat the water in this tank from $23 ^ { \circ } \mathrm { C } \text { to } 75 ^ { \circ } \mathrm { C } \text { ? }$ The specific heat of water is $4186 \mathrm {~J} / \mathrm { kg } \cdot \mathrm { K }$ and its density is $1000 \mathrm {~kg} / \mathrm { m } ^ { 3 }$

Question

A 200-L electric water heater uses  2.0 kW. Assuming no heat loss, how many hours would it take to heat the water in this tank from $23 ^ { \circ } \mathrm { C } \text { to } 75 ^ { \circ } \mathrm { C } \text { ? }$ The specific heat of water is $4186 \mathrm {~J} / \mathrm { kg } \cdot \mathrm { K }$ and its density is $1000 \mathrm {~kg} / \mathrm { m } ^ { 3 }$

Quizplus · Accepted Answer

The energy required to heat the water is:$$Q = mc_p \Delta T = (200 \mathrm{~kg})(4186 \mathrm{~J} / \mathrm{~kg} \cdot \mathrm{~K})(75^{\circ} \mathrm{C} - 23^{\circ} \mathrm{C}) = 4.35 	imes 10^7 \mathrm{~J}$$The power of the water heater is:$$P = 2.0 \mathrm{~kW} = 2000 \mathrm{~W} = 2000 \mathrm{~J} / \mathrm{s}$$Therefore, the time required to heat the water is:$$t = \frac{Q}{P} = \frac{4.35 	imes 10^7 \mathrm{~J}}{2000 \mathrm{~J} / \mathrm{s}} = 21750 \mathrm{~s} = 6.04 \mathrm{~hours}$$

A 200-L Electric Water Heater Uses 2 23∘C to 75∘C ? 23 ^ { \circ } \mathrm { C } \text { to } 75 ^ { \circ } \mathrm { C } \text { ? }23∘C to 75∘C ?

A 200-L Electric Water Heater Uses 2 $23 ^ { \circ } \mathrm { C } \text { to } 75 ^ { \circ } \mathrm { C } \text { ? }$