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Question: An electric heater of resistance \(42 \Omega\) converts 400 g of ice to water at 70°C in 4 minutes. ...

An electric heater of resistance 42Ω42 \Omega converts 400 g of ice to water at 70°C in 4 minutes. Calculate the current drawn by the heater. [Specific heat capacity of water= 4200Jkg1K14200 J{kg}^{-1}{K}^{-1}, Specific latent heat of ice= 336000Jkg1336000 J{kg}^{-1}].

Explanation

Solution

To solve this problem first find the heat energy. Use the formula for specific heat and find the amount of heat energy. Then, substitute this value in the formula for power giving relationship between heat transfer and time. We also know the formula for power in terms of current and resistance. So, substitute the value of power obtained and the value of resistance given. This will give the amount of current drawn by the heater.

Formula used:
Q=mL+mSΔTQ = mL + mS \Delta T
P=QtP= \dfrac {Q}{t}
P=I2RP= {I}^{2}R

Complete answer:
Given: Resistance of heater (R)=42Ω42 \Omega
Time (t) = 4 minutes= 240 secs
Mass of ice (m)= 400 g= 0.4 kg
Specific latent heat of ice (L)= 336000Jkg1336000 J{kg}^{-1}
Specific heat of water (S)= 4200Jkg1K14200 J{kg}^{-1}{K}^{-1}
Initial temperature T1=70°C{T}_{1}= 70°C
Final temperature T2=0°C{T}_{2}= 0°C

Specific heat formula is given by,
Q=mL+mSΔTQ = mL + mS \Delta T
Q=mL+mS(T1T2)\Rightarrow Q= mL + mS({T}_{1}-{T}_{2})
Substituting the values in above equation we get,
Q=(0.4×336000)+(0.4×4200×(700))Q= (0.4 \times 336000)+ (0.4 \times 4200 \times (70-0))
Q=134400+117600\Rightarrow Q = 134400 + 117600
Q=252000\Rightarrow Q= 252000
We know, Power is given by,
P=QtP= \dfrac {Q}{t}
Substituting the values in above equation we get,
P=252000240P = \dfrac {252000}{240}
P=1050Watt\Rightarrow P= 1050 Watt
Relationship between power and current is given by,
P=I2RP= {I}^{2}R
Rearranging the above equation we get,
I2=PR{I}^{2}= \dfrac {P}{R}
Substituting the values in above equation we get,
I2=105042{I}^{2}=\dfrac {1050}{42}
I2=25\Rightarrow {I}_{2}= 25
Taking the square root on both the sides we get,
I=5AI= 5 A
Hence, the current drawn by the heater is 5 A.

Note:
Before solving the problem, students should look at the units whether those are balanced or not. If the units are not mentioned then convert them before applying them. In this question, we have obtained the value of heat energy so we should have an idea about what is heat energy? Heat energy is the energy which is due to small particles like atoms, molecules or the ions in solids, liquids or gases. Transfer of heat energy is possible by three methods. Those methods are Conduction, Convection and Radiation.