Final answer:
To convert 74 grams of ice at -4°C into 74 grams of water at 52°C, we need to calculate the heat required for temperature change and phase change. First, we heat the ice from -4°C to 0°C, then calculate the heat required to convert the ice at 0°C to water at 0°C, and finally heat the water from 0°C to 52°C. The total heat required is 41341.36 J.
Step-by-step explanation:
To calculate the heat required to convert 74 grams of ice at -4°C into 74 grams of water at 52°C, we need to consider the temperature change and the phase change. First, we need to heat the ice from -4°C to its melting point of 0°C. The heat required can be calculated using the equation Q = mCpΔT, where Q is the heat, m is the mass, Cp is the specific heat capacity and ΔT is the change in temperature.
Q = 74 g x 2.06 J/g°C x (0°C - (-4°C)) = 607.04 J
Next, we need to calculate the heat required to convert the ice at 0°C into water at 0°C. This is the heat of fusion, which is 334 J/g. Therefore, the heat required is given by:
Q = 74 g x 334 J/g = 24716 J
Lastly, we need to heat the water from 0°C to 52°C. Using the same equation as before:
Q = 74 g x 4.18 J/g°C x (52°C - 0°C) = 16018.32 J
To find the total heat required, we sum up the heat calculated above: 607.04 J + 24716 J + 16018.32 J = 41341.36 J.