Final answer:
To take a 49.3-gram block of ice from -22.2°C to 22.2°C, the total heat required is 23304.678 joules or 5569.113 calories, factoring in the heat required to warm the ice, melt it, and then heat the water.
Step-by-step explanation:
To calculate the total amount of heat needed to take a 49.3-gram block of ice from -22.2°C to 22.2°C, one must consider the heat required to raise the temperature of ice to 0°C, the heat of fusion to melt the ice into water, and then the heat required to raise the temperature of that water to 22.2°C. The specific heat capacity of ice is approximately 2.09 J/g°C and the specific heat capacity of water is 4.18 J/g°C.
First, calculate the heat required to warm the ice from -22.2°C to 0°C using the equation Q = mcΔT, where 'm' is the mass, 'c' is the specific heat capacity, and 'ΔT' is the change in temperature. Then add the heat of fusion, which is required to turn the ice into water: Q = m x heat of fusion. Finally, calculate the heat required to increase the temperature of the resulting water from 0°C to 22.2°C.
Heat required to warm the ice = m * c * ΔT = 49.3 g * 2.09 J/g°C * (0 - (-22.2))°C = 49.3 g * 2.09 J/g°C * 22.2°C = 2277.274 J
Heat required to melt the ice = m x heat of fusion = 49.3 g * 334 J/g = 16473.2 J
Heat required to heat the water = m * c * ΔT = 49.3 g * 4.18 J/g°C * 22.2°C = 4554.204 J
Total heat required = Heat required to warm the ice + Heat required to melt the ice + Heat required to heat the water = 2277.274 J + 16473.2 J + 4554.204 J = 23304.678 J
To convert this energy to calories, we use the conversion factor 1 calorie = 4.184 joules.
Total heat required in calories = 23304.678 J / 4.184 J/cal = 5569.113 calories
Therefore, the total heat required to take the 49.3-gram block of ice from -22.2°C to 22.2°C is 23304.678 J or 5569.113 calories.