Answer:
66 grams of ice would have to melt to lower the temperature of 352 mL of water from 15 °C to 0 °C.
Step-by-step explanation:
To calculate the amount of ice that would have to melt to lower the temperature of 352 mL of water from 15 °C to 0 °C, we need to use the formula:
Q = m_water * c_water * ΔT_water + m_ice * Lf
where,
Q = the amount of heat transferred,
m_water = the mass of water, c_water is the specific heat capacity of water,
ΔT_water = the change in temperature of water, m_ice = the mass of ice,
Lf = the specific latent heat of fusion of ice.
First, let's calculate the amount of heat transferred to the water:
Q = m_water * c_water * ΔT_water
Q = 352 g * 1.0 cal/(g*°C) * (15-0) °C
Q = 5,280 cal
Next, we can use the specific latent heat of fusion of ice, which is 80 cal/g, to calculate the amount of heat required to melt the ice:
Q = m_ice * Lf
Q = m_ice * 80 cal/g
m_ice = Q / Lf
m_ice = 5,280 cal / 80 cal/g
m_ice = 66 g