Sebanyak 200 g air dengan suhu \( 30^{\circ} \mathrm{C} \) dicampurkan ke dalam 100 g air panas dengan suhu \( 90^{\circ} \mathrm{C} \). Jika kalor jenis air adalah \( 1 \mathrm{kal} / \mathrm{g}{ }^{\circ} \mathrm{C} \), suhu air setelah dicampurkan adalah .... \( 50^{\circ} \mathrm{C} \) \( 75^{\circ} \mathrm{C} \) \( 100^{\circ} \mathrm{C} \) \( 125^{\circ} \mathrm{C} \) \( 150^{\circ} \mathrm{C} \) Pertanyaan ini wajib diisi

Let the mass of the cool water be \( m_1 = 200 \, \mathrm{g} \) at an initial temperature \( t_1 = 30^{\circ} \mathrm{C} \) and the mass of the hot water be \( m_2 = 100 \, \mathrm{g} \) at an initial temperature \( t_2 = 90^{\circ} \mathrm{C} \). Let the final equilibrium temperature be \( t \). Since no heat is lost to the surroundings, the heat gained by the cooler water is equal to the heat lost by the hotter water: \[ m_1 \, c \, (t - t_1) = m_2 \, c \, (t_2 - t) \] Given that the specific heat capacity \( c = 1 \, \mathrm{kal/g^{\circ}C} \), we can simplify the equation: \[ 200 \, (t - 30) = 100 \, (90 - t) \] Now, solving for \( t \): 1. Expand both sides: \[ 200t - 6000 = 9000 - 100t \] 2. Add \( 100t \) to both sides: \[ 200t + 100t - 6000 = 9000 \] \[ 300t - 6000 = 9000 \] 3. Add \( 6000 \) to both sides: \[ 300t = 15000 \] 4. Divide both sides by \( 300 \): \[ t = \frac{15000}{300} = 50 \] Thus, the final temperature is: \[ t = 50^{\circ} \mathrm{C} \]