How much ice at 273 K can be melted by 5.00 * 10^3 J? ?

The specific heat capacity of water is about 4.18.

That means it takes 4.18 joules to increase the temperature of 1 gram of water by 1 degree Celsius (or 1 Kelvin).

In this case, we need to increase the temperature of the ice to the melting point of water, which is 0 degrees Celsius, or approximately 273.15 Kelvins; so we need to increase its temperature by 273.15 - 273 = 0.15 Kelvins or 0.15 degrees Celsius.

So for that, we need 4.18 * 0.15 = 0.627 joules per gram.

If you have 5.00 * 10^3 = 5000 joules, divide that by 0.627 to get approximately 7974.4816586921850079744816586922 grams; almost 8 kilograms.

That's around 8 litres of ice.

(5000 J)/(334 J/g) = about 16 grams but use a calculator.