Question

A(n) 70-g ice cube at 0°C is placed in 960 g of water at 24°C. What is the final temperature of the mixture?
______ °C

Answer 1

we know that

heat lost by the water is equal to the heat gained by the ice from the principle of calorimetry

Q_{loss by water} = Q _{heat gained by ice}

m_wc_w ( 24^o - T ) = m_ice L_f + m_icec_w (T - 0^o C )

m_wc_w ( 24^o - T ) = m_ice L_f + m_icec_w T

m_wc_w ( 24^o) - m_wc_w T = m_ice L_f + m_icec_w T

m_icec_w T + m_wc_w T = m_ice L_f - m_wc_w ( 24^o)

T = ( m_wc_w ( 24^o)- m_ice L_f  ) / ( m_icec_w + m_wc_w )

given data

m_ice=70 g = 70x10^-3 kg

m_water= 960 g = 0.96 kg

c_w= 4.18x10³ J/kg^oc

L_f= 333.5x10³ J/ kg

plugging the values in the formula

T = ((0.96)*(4.18x10³)*(24)-(70x10^-3)*(333.5x10³)) / ((0.96)*(4.18x10³) + (70x10^-3)*(4.18x10³) )

T =16.95^oC

final temperature of the mixture = T = 16.95^oC