Question

Can someone solve this question for me​

Answer 1

The answer is approximately 1100 cal

While we convert ice to water, we have to convert ice into water at 0°C then water at 0°C to water at 30°C

1) So ice to water

q = mL

Latent heat = 334 J/g

q= 100 × 334

q= 33400 J

2) From water at 0° to water at 30°C

q= mc∆T

q = 100 × 4.18 × (30-0). { c = 4.18J/C}

q= 12558 J

Total energy required is

qtotal = 33400+12558

q = 45958J

So when we convert it into calorie we will divide it by 4.18

Then,

q = 45958/4.18

q≈ 1100 cal

( But real answer is 10994)

We have done approximation

Hope it helped you dear friend