Question

A tub containing 53 kg of water is placed in a farmer's canning cellar, initially at 10∘C. On a cold evening the cellar loses thermal energy through the walls at a rate of 1200 J/s. Without the tub of water, the fruit would freeze in 4 h (the fruit freezes at −1∘C because the sugar in the fruit lowers the freezing temperature). Part A By what time interval does the presence of the water delay the freezing of the fruit? The specific heat of water is c = 4180 J/kg⋅∘C.

Answer 1

To solve this problem it is necessary to apply the concepts related to heat transfer and power depending on energy and time.

By definition we know that the heat loss of water is given by

$Q = mc_w*\Delta T+m*L_f$

Where,

m = mass

$c_w =$ Specific Heat of Water

T = Temperature

$L_f =$Latent heat of fusion $\rightarrow$ Heat of fusion for water at 0°C is $3.35*10^5J/Kg$

Our values are given as,

m=53 kg

$C=4180 J/kg\°C$

$\Delta T=10-(-1)=11$

Replacing we have,

$Q=53*4180*6+53*3.35*10^5$

$Q = 19084240J$

Power can be defined as

$P = \frac{Q}{t}$

Re-arrange to find t,

$t = \frac{Q}{P}$

$t = \frac{19084240}{1200}$

$t = 15903.53s \approx 265 min \aprox 4.41h$

Therefore the time interval is 4.41h