Final answer:
Approximately 17.3 grams of ice is required to cool 60 g of water from 43 °C to 20 °C, by using the specific heat capacity of water and the heat of fusion of ice.
Step-by-step explanation:
The mass of ice needed to cool 60 g of water from 43 °C to 20 °C can be calculated by using the principle of conservation of energy, which states that the heat lost by the water must be equal to the heat gained by the ice. To find this, we use the specific heat capacity of water (4.184 J/g°C) to calculate the heat lost by the water when it cools, and the enthalpy of fusion of ice (334 J/g) to calculate the heat needed to melt the ice.
The heat lost by water (q) is calculated as follows:
q = m * c * ΔT
where m is the mass of the water, c is the specific heat capacity, and ΔT is the change in temperature. Thus:
q = 60 g * 4.184 J/g°C * (43 °C - 20 °C) = 60 g * 4.184 J/g°C * 23 °C = 5779.52 J
The mass of ice needed to absorb this heat (m_ice) when it melts can then be calculated using the enthalpy of fusion:
m_ice = q / enthalpy of fusion
m_ice = 5779.52 J / 334 J/g ≈ 17.3 g
Therefore, approximately 17.3 grams of ice is required to cool 60 g of water from 43 °C to 20 °C.