Question

A refrigerator door is $150\text{cm}$ high, $80\text{cm}$ wide and $6\text{cm}$ thick. If the coefficient of conductivity is $0.0005{\text{cal/cm-s-}}^{\circ }\text{C}$ and the inner and outer surfaces are at $0^{\circ }\text{C}$ and ${30}^{\circ }\text{C}$ respectively, find the heat loss per minute through the door. (in Calories)

• A. $3600$
• B. $1800$
• C. $900$
• D. $1200$

Answer 1

The correct option is B $1800$

Given,
Height of refrigerator $\left(h\right)=150\text{cm}$
Width of refrigerator $\left(w\right)=80\text{cm}$
Thickness of refrigerator $\left(d\right)=6\text{cm}$
Coefficient of thermal conductivity $\left(k\right)=0.0005{\text{cal/cm-s-}}^{\circ }\text{C}=0.0005\times 60{\text{cal/cm-min-}}^{\circ }\text{C}$
Temperature difference between inner and outer surfaces $(t_h-t_c)={30}^{\circ }\text{C}$

Applying the equation of thermal conductivity,
$\frac{Q}{t}=\frac{kA(t_h-t_c)}{d}$
For t=1 minute:
$\Rightarrow Q=\frac{0.0005\times \left(60\right)\times (150\times 80)\times \left(30\right)}{6}\Rightarrow Q=1800\text{cal/min}$