Question

What is the rate of heat transfer required to melt 1-ton of ice at 32 F in 24 hours?

Answer 1

Answer:

3865.74 J/s

Step-by-step explanation:

mass of ice, m = 1 ton = 1000 kg

time , t = 24 hours

latent heat of fusion of ice, L  = 334000 J/kg

Heat required to melt, H = m x L

where, m is the mass of ice and L be the latent heat of fusion

So, H = 1000 x 334000 = 334 xx 10^6 J

Rate of heat transfer = heat / time = $\frac{334\times 10^{6}}{86400}$

Rate of heat transfer = 3865.74 J/s

thus, the rate of heat transfer is 3865.74 J/s.