Depends on a few things, but...a half inch maybe? It takes time. And growth occurs slower roughly exponentially as a function of thickness, so if it is already an inch thick then it takes a lot longer to add another half inch than it would for water to initially freeze a half inch.
deformation
(-5) degrees Celsius = 23 degrees Fahrenheit.
5 degrees is colder
5 degrees Celsius is 41 degrees Fahrenheit.
5 degrees Celsius = 41 degrees Fahrenheit.
Ice
5 more degrees
no it melts at 0 degrees Celsius or higher.
Ice cream is typically frozen at temperatures below 0 degrees Celsius (32 degrees Fahrenheit). Most household freezers are set to a temperature around -18 degrees Celsius (0 degrees Fahrenheit) to properly freeze ice cream.
deformation
Ice cream needs to be kept below freezing, because it has ICE in it, and ice needs to be kept at 0 degrees centigrade or lower. On average a fridge is 4 degrees centigrade so it will melt in the fridge...
It depends on what "5 thick" is: 5 inches thick, 5 feet thick, 5 yards thick, 5 miles thick etc.
If the temperature of the earth went up by 5 degrees, the only place polar bears would survive is in a zoo. Polar bears need the ice because they feed through holes in the ice. If there is no ice, then they can't feed.
You never drink either (100°C is boiling, or steam, and -5 °C is normally ice).
One morning, the temperature was -5˚F. By noon, the temperature had increased 12 degrees
A block of ice with the dimensions listed has a volume of 35ft^3. The density of ice at 32F is about 57.24 lb/ft^3. Multiplying the two together gives a mass of the ice of about 2003.4 lbs.
Heat required for this transition is given as the the sum of three heatsheat required for heating the ice from -5 degree Celsius +latent heat(conversion of ice at zero degree to water at zero degrees)+heat required to heat the water from 0 to 5 degree CelsiusHeating of ice=m x s x delta T,where m is the mass ,s is the specific heat of ice=200x0.5x5=500calmelting of ice=mxlatent heat=200x80=16,000calHeating of water=m x s x delta T,where m is the mass ,s is the specific heat of water =200x1x5=1000calTotal heat required=500+16,000+1000=17,500 cal